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C7.

LJ

r

- yy t

s

«^,



WORKS

BY

THE

SAME

AUTHOR

Published

by McGRAW-HILL BOOK COMPANY

The

Electric Circuit

XV

+

229 pages, 6

by 9.

Cloth

$2.00

(Second

Edition, entirely rewritten

and

enlarged)

The

Magnetic Circuit

xviii

+

283 pages, 6

by 9.

Cloth

$2.00

Published

by

JOHN

WILEY

&

SONS,

Inc.

Experimental Electrical

Engineering

Vol. I.

xix

+

469

pages,

6 by

9,

328 figures.

Cloth $3.60

net

Vol. II.

xiv

+

333

pages, 6

by

9,

209

figures.

Cloth

$2.50 net

Engineering

Applications

of

Higher

Math-

ematics

Part

I.

Machine

Design,

xiv

+

69

pages,

5iby8.

Cloth $0.75 net

Part

II.

Hydraulics,

v

+

103

pages,

5J

by

8.

Cloth

$0.75

net

Part

III.

Thermodynamics, v

+

113

pages,

5i

by

8.

Cloth $0.75

net

Part

IV.

Mechanics

of Materials.

V

+

81

pages,

5|

by

8.

Cloth

$0.75 net

Part

V.

Electrical

Engineering,

vii

+

65

pages,

5i

by

8.

Cloth $0.75

net

Elementary

Electric

Testing

Loose

Leaf

Laboratory

Manual

of the

Wiley

Technical

Series,

J.

M. Jameson,

Editor.

25

direction

sheets

with numerous

diagrams

and

cuts.

Complete

in removable

paper

cover

$0.50

net

Published

by

FERDINAND

ENKE,

STUTTGART

Ueber

Mehrphasige

Stromsysteme

bei

.

Ungleichmassiger

Belastung.

Paper.

. .

.

Mk. 2.40



ENGINEEHINii

APIM.K'ATIONS

OK

TGHER

MATIIKMATICS

BY

V.

KAKAPCTOFF

PART

I.

PROBLEMS ON

MACHINE

DESIGN

FIRST

EDITION

SBCOND

IMPRESSION,

CORRRCTED

NEW

YORK

JOHN

WILEY

&

SONS

Lomooh:

chapman

k HALL.

Limitbu

1917



COPTRIOHT,

1911,

BY

V.

KARAPETOFF

330

pi-

Stanbopc

lpree&

F. H.

GILSON

COMPAMT

BOSTON.

U.S.A.



PREFACE.

This

is

not

a

book

on

calculus

or

analytic

geometry

(the

market

is flooded

v^ith

them);

nor

is

this

a

book

on

engineering

or any

branch of

it.

The book

is

intended

to

enable an

engineer to

make a

belter

and

more

extended

tue

of

higher

mathematics

in

his work.

The

purpose of

the

book

may

be best

ampUhed by a

parable.

In a

manual-training

school

(on

the

moon)

machinist

apprentices

were

taught

their

trade

in

the

following

manner: During

the

first

year

they

had

a

highly

theoretical course

on

the subject of

various

tools

used

on

lathes,

planers, boring

mills, milling

machines,

etc.

The shapes of

the

tools were derived

and

ex-

plained

in

detail

on comphcated

drawings;

most

gen-

eral

theorems were proved

concerning

these

tools;

it

was

shown

how

to design these

tools,

not

only for

a

few

simple

practical cases,

but principally

for many

hypothetical cases which were supposed to be

of

some

importance

on

Mars.

This latter part of the

course

was

justified

on

the

plea

of

mental

gjinnastics. No

actual machine-tools

were

provided

in this department

and

no

practice

was afforded the student

in

the

use

of

the

tools.

During

the next

two

years

the students

were required

to

finish, fit,

and assemble

the

parts

of

various

engines

and

other

pieces of

mechanical

apparatus.

Had

t)u

\

l^een

previously

trained in the

use

of

machine-tools,

their

shop-work

would

have been much

simpUfied. But

their

highly

theoretical

information

about

tools

was

of

no

iU



«

IV

PREFACE.

use

to them,

even

if

they

had not forgotten

it during

the

summer

vacation.

At

any

rate,

they

did

not

pos-

sess

enough manual

skill

to

put

even

an

iron rod

in

a

lathe

and

turn it

down

by a

millimeter

or

two.

So

they

preferred to

use

the

old-fashioned

chisel

and

file,

and

finished

most of their

parts

in

a

vise.

And

their

teachers

acquiesced

in this way

of

doing

things,

partly

because they

did not

consider it

to be

their

duty

to

teach the use

of

tools,

since

that was

a part

of

the

freshman-year

instruction

and was

taught

by

another

department,

and

partly because,

having

gone themselves

through

a

similar

school,

they

considered such

a

lack

of

correlation

in teaching to be

inevitable.

Some

teachers

even

considered

this state of

affairs

to

be

a

fundamental

law

of

nature,

second

only

to

the

law

of

gravitation.

But

a

heretic,

an

anarchist,

an iconoclast, is

sure to

appear

on the stage

sooner

or

later.

Such

a

one made

his

appearance

one

day

and began undermining

the

pillars

of the

time-honored system.

He

claimed

that

it

was

wrong

to

instruct

freshmen

in

the

theory

of the

tools,

without

having

applications

and practical shop

work

going

hand

in

hand.

His

idea

was to

show the

student

one

or

two

simple tools

and

to

apply

them

im-

mediately

to

finishing

a

few

pieces

of

work

in

a

suitable

machine-tool.

In other

words,

according

to

this

man,

knowledge

and

skill,

or science

and

art, ought to

be ac-

quired

simultaneously.

The

reformer

insisted

that

with

this

method

of instruction

the knowledge

gained

would

become

organic

with

the

student, instead

of

being

on

the

surface

only.

Besides,

said

he,

modern

psychology

shows

that

interest

is a

paramount

factor

in

education,

and

applications are

always

more

interesting

to

an

average

student

than

a

general

theory.

He

was

per-



PRBFACB.

fectly

willing

to

grant

that

with his method,

consider-

ably

less

ground

could

be

covered

in

pure

theory during

the

first

year;

but

then he

claimed that

machinist

prentices

needed

only a

few

simple tools

during

the

t

year's

work, and

that

enough

time could be spared

later

on to

give

an

advanced

course in the theory

of

special

and

fancy

tools

to

those

who

desired it.

This

time

could

be

spared, because the

practical knowledge

and

skill

in

the

use

of

tools,

acquired

during the first

year,

would enable the students to

accomplish their

shop-work

in

less time.

WTiile

it may

seem

incredible

that mechanic

arts

should

be

taught anywhere in the

ridiculous

way

described

above,

yet

it is a

fact that mathematics is

most

universally

taught

to

engineering

students

in

a

similar

fashion,

without any correlation whatever

with their profession. The

student

is

first

filled

with

analytics

and

calculus,

as

if he

expected to

become

a

pure

mathematician.

Then,

very little of

this

mathe-

matics is

used

in

the engineering courses, partly

be-

cause

the students

find

it very difficult

to follow,

and

partly

because

many

professors

of

engineering

them-

selves

have

not

enough

grounding

in mathematics

to

feel

at ease in it

and

to

make

it interesting

and

attrac-

tive

to their

students. The case

is

somewhat similar

to

that

of

modem

languages.

The engineering

facul-

ties

insist

that

students

shall

acquire

proficiency

in

French

and

German,

while the

students

know

perfectly

well

that

most

of

their

teachers never read foreign

books or

magazines.

The

old saying about

the

mote

and

the

beam

in the eye

involuntarily

suggests

itself.

The

heretic

who

first raised his voice

against

the

unsatisfactory teaching

of mathematics

to engineering

students,

and

who

showed

the

way

out

of

it,

was

the



VI

PREFACE.

noted British

engineer

and

educator,

John

Perry. In

his

Calculus for

Engineers

(1897),

he limits the

theory

to

a

minimum,

and

gives a large number

of

examples

taken from various

branches of engineering.

After

the student

has

mastered the differentiation

and

the

integration of

a

few

principal functions

and has

solved

a

large number of practical

engineering

problems,

involving

these

functions, he

is

led into a

more general

theory

of differentiation

and

integration,

somewhat

as

it

is

done

in

the

ordinary

courses

in

calculus.*

Perry's pioneer work,

and

the so-called

Perry

Movement

for

teaching mathematics in

a

practical

way, will

forever

remain

important mpnuments

in

the

history

of engineering

education.

Like

many

other pio-

neer works.

Perry's

Calculus

proved

to

possess

some

de-

fects

when used as

a

regular textbook,

and

several

similar

works

by

other writers

followed

it (see Appendix).

This book differs from

that

of

Perry

and his followers

in

two

respects:

(1)

An

elementary knowledge

of analytics

and

cal-

culus

is

presupposed;

(2)

The arrangement of

the

chapters is according

to

the

engineering

topics

and not

according

to

the

mathe-

matical

functions

or operations.

It is

not necessary

to

burden a

book

of this charac-

ter

with the theory

of analytic

geometry

and

calculus,

because

it is

treated

in numerous

excellent

works,

large

and

small.

The

names

of

some

of

these

books

are

given

in

the list

at the end

of

this

book. This

book

*

For a detailed development of

this

idea

in

application

to ill engi-

neering subjects, see

V.

Karapetoflf, On

the

Concentric

Method of

Teaching

Electrical

Engineering, in

Trans. Amer.

Inst.

Electr. Engrs.,

Vol.

26

(1907),

p.

1441; also

his

paper

On the

Concentric Method

of

Education

in

Engineering,

Proc.

Soc. Promotion

Eng. Educ., Vol.

16

(1908),

p.

258.



r

PREFACE.

vii

is intended to

supply the

engineering applications

and

reasoning not

found

in the

other books.

The

arrange-

ment

by

engineering

topics

permits

the

gi\ing of

more

interesting

examples

than if they

were arranged

accord-

ing to the

mathematical functions or

operations;

namely,

having

explained

to

the

student the

purpose

of a simple

engineering

appliance, such

as

a

belt or

a

flywheel,

the

teacher

can

give

a

number

of

mathematical

examples

on

these appUances,

from

the

most

elementary

to

the

most

difficult,

without going

into

engineering

practice.

In

other

words,

the

aim

is

to enable

a

teacher

of

math-

ematics

to

give

a large

number

of

practical

problems,

with

verj' little

engineering

information.

All

the nec-

essary

explanation in regard

to

the construction

and

use

of

each

device

is

given

at

the

beginning

of

each

chap-

ter, so

that

it

is

not

necessary to

look up any

books

on

macliine

design.

However,

a

list

of such

engineer-

ing

works will

be

found

at

the

end

of

this

book.

In

accordance

with

the above-described

purpose and

plan

of

the

book,

the following possible

uses

of

it

are

suggested:

(a)

As

a problem book

in connection with

a

regular

course in

analytics

and

calculus.

(6)

As a

textbook

in

a supplementary

course

(after

the

completion

of

a

brief

course in calculus)

taught

in

the

department of

mathematics for

the purpose of

fixing

the

mathematical

operations

in the mind

of

tlic

student

and

preparing

him

for

the

enginc»ering subjects.

(c)

As

a

textbook

in a course hi

Enginc^'ring

Mathe-

matics,

taught

in

the

junior

or

in

the senior

year

by

an

engineer,

to

illustrate

the methods

of

engineering

research

and

analysis.

(«0

As

a

text-

or

reference-book in

a

seminar

for

grad-

uate

students in engineering. Most of

these

men

come

h



VIU

PREFACE.

to colleges

with

a

knowledge of

calculus

which

is

anything but

good,

and

consequently are handicapped

in

their

research and

in understanding

the literature

of

the subject

upon

which

they are

working.

(e)

As a study book

for

teachers

in

engineering

and

for

practicing

engineers,

who require mathematics

in

their

work and

feel

that

they

need

a

brushing

up

in

order

to

be

able

to follow intelUgently books,

maga-

zines,

and

transactions

of

engineering

societies.

While

the

problems

in this

volume have

been

selected

from

the theory of

machine elements,

the

book

might

be

of

interest to engineers of

different

kinds. It

is

the

reasoning

and the

practice

in

the

use

of

calculus

and

analytics that is

of

importance, and

not the

particular

kind

of

appliances

to

which

the

problems

refer.

However, the

author

has also

collected

for his

students a

large mmiber

of

problems

taken

from

HydrauUcs,

Strength

of

Materials,

Thermodynamics,

and

Electrical

Engineering. It is his intention

to pub-

Hsh

these

problems in similar

small

volumes.*

The author wishes

to

acknowledge

with

thanks

the

moral

support, encouragement,

and

assistance on

the

part

of some members of

the

mathematical

faculty

of

Cornell

University

—

in

particular.

Professors James

McMahon and F.

R.

Sharpe.

Mr.

John G. Pertsch,

Jr.,

instructor

in electrical

engineering

in

Sibley

College,

read

the

manuscript

and

the proofs

of

the

book

and

checked

the

solutions of

the

problems.

To

him

the

author

is under

obhgation

for

the

care

with which

the

work

was done.

Cornell University,

Ithaoa,

N.

Y.

September,

1911.

Revised April,

1917.

*

These

four

volumes

were published in

1916.



A

DIALOGUE

Brwcbn the

PMepscnvc

Utn

or thb

Book and

tbb Aotbok.

Prospective

Uaer.

—

You assume

in your

book

that

I

can

differentiate

and

integrate

almost any function,

but such is

not

the case.

Author.

—

I

assume

only

that

sometime

you

have

studied

the

elements of

calculus, and

that by

referring

to your old

booLs

you

can

refresh

your

memory

as

to

the

processes of

differentiation

and

integration.

The

specific

information, such as formulae,

routine transfor-

mations, etc.,

you

will

find

in the

reference

books by

Seaver, Claudel,

and

Peirce,

mentioned in the Appen-

dix.

You

should have at

least

Seaver's

book

for

reference,

while

studying

my

book; this

will

save

you

considerable

time.

P.

U.

— Yes,

but even

where

you

indicate

the

solution you

seem to omit

important steps

in

trans-

formations;

at least

so it

appears

to

me

because

your

deductions

and

prrM)fs

are

so

short.

A.

—

Engineering

problems

and

deductions must

be short, because the

assumptions

made

in the

begin-

ning

are

somewhat

crude

an>'Avay. In solutions

I omit

only

such

stojw as

are taken

according

to

the well-

known

rules

of elementary

or higher mathematics.

The

student

has no

choice

there

and

is

sure

to fill

in

the

omitted

parts correctly. If I

should

give the

details

of

t ranformations, the principal reasoning would

be

obscured.

ix



X

A

DIALOGUE.

P'

U.

—

This may

be

so,

but I

had

my

algebra

and

trigonometry

so long

ago

that

I

am

not

certain

even

about

the

simplest

transformations

and

relations,

and

I

should

much

rather

have

them

indicated

in

your

book.

A.

—

No

one

remembers

all

the

formulae,

relations,

and

transformations,

not

even

a

professional

mathe-

matician.

I

want

you

to

go back to

your

old

books

and

to

the

reference

books

of

which

I

spoke

before,

and

get

the specific

information

which

you

need

in

each

problem. In

this

way

you will

learn

much more

and

will remember

better than if

I

gave

you

all

the

steps.

The

profit

derived

is

in

proportion

to

the indi-

vidual

effort

made

and

not according

to

the

number

of

pages

gone

over

in

a

perfunctory

way.

P.

U.

—

I

am

willing to

grant

you

this

point,

but

I want

you

to

understand

that

I am

no

exception

as

far

as the

knowledge of elementary

mathematics is

concerned.

I

think

my

knowledge

of it

was

above

that

of an

average

student, but I

never used

enough

of

algebra or

trigonometry

to

make my knowledge

organic.

Do

you

not

think

that

we

ought

to

have,

in

the

freshman

year,

problems

on

engineering appli-

cations of elementary

mathematics,

before

we

embark

on analytic geometry

at all? My principal

difficulty

with

higher mathematics

was that

I could not

follow

the algebraic

transformations

fast enough,

and there-

fore

lost

the

principal

thought.

I

always

felt

like

a

man

who

came

to the theater in the

middle

of the sec-

ond

act, and

tried

to

follow

the

plot of a

comphcated

play.

A,

—

I fully

agree

with

you

there.

Colleges of

engineering ought

to

see

to

it

that the

course

in mathe-

matics begins

with

a

series

of

practical

problems in



A

DIAlJOaVE.

n

elementary mathematics,

before anal3rtics or

calculus

are

touched

at

all.

Saxelby's

book,

mentioned

in

the

Appendix,

is ver\' good

for

the purpose.

I

trust that

someone

will

write a

small

book on

the

engineering

applications

of

clementar>'

mathematics. As

for

you,

I

am

certain that you will

incidentally

acquire

a con-

siderable

fluency

in algebra

and

trigonometry

by

wav-

ing the

priiblems

in my

lKx>k.

P.

U.

—

You

do

not

give

much numerical work in

your

book.

Do

you

not think

that

fluency

in

com-

putations

is

quite essential

for

the

engineer?

A.

—

I do,

and

I

require numerical computations

in

those

problems

where such

computations

or

the numer-

ical

results

bring

out

a new point.

To

do

more

than

this

would

distract

attention

from

the

principal

aim

of

the book.

This

aim,

as

you

understand,

is to train

the

engineering student

in such

reasoning

in his

pro-

fession

where

higher

mathematics

can

be used

to ad-

vantage.

I even

avoided giving

numerical

coefficients

wherever

possible,

because such are given

in engineer-

ing pocket books

and handbooks.

Please do not

as-

cribe much importance to

the numerical

data in the

problems.

They are

given

merely

for exercise.

/'

U.

—

You

have

just

mentioned

engineering reason-

ing.

Is it

any

different from

that used

in

other sciences?

A.

—

I

should

have

said reasoning

in

application to

engineering

problems.

It means first

to size

up the

problem

in

regard

to

the

physical

laws

involved

and

the empirical

assumptions to

be

made.

After this,

the

problem

is

to

be

clothed in

a mathematical

lan-

guage.

Then comes

the

solution,

with the necessary

transformations,

approximations,

etc.

The

result

must

be

analyzed

as

to

its

applicability to

the

particular

practical

problem.

This

means

going

back

to

the

k



xii

A

DIALOGUE.

physical laws

and

to the engineering

assumptions

made.

It is

always advisable

to

apply the

solution

to

at

least

one set of

numerical

data, or, still

better,

to

some average

data

and to

extreme

data.

The numer-

ical results often

suggest a

new solution, different

assumptions,

another

approximation,

or at least

a

different

form of

the

solution.

P.

U.

—

What you have just enumerated

really

requires

a

man

to

be

a

competent

engineer,

and

at

the

same time

a

first-class mathematician. This is a rare

combination. Cannot

two

men,

an

engineer

and

a

mathematician, solve

such

problems

together,

in

a

cooperative way?

A.

—

Yes,

they

could if they understood each other

better

than

they

do

now.

They

could

if

the engineer

were

able to

put

his

problem into

the

mathematical

language,

stripping it

of

all

the

descriptive

factors

which

cannot

be

expressed

by

formulae;

and

if

the

mathematician had

a

little

engineering feeling

which

would enable

him

to suggest

more

suitable

fundamental

assumptions,

and to present the

solution

in

a

form

which

would

not

scare a practician by

its

very

size.

My book is

intended

to

bring these

two

classes

of

men into a

closer

relation.

Suppose that you

were

in

France

and

had

to

write

a

letter

in French. If

you

knew

the

French language

well

enough to

explain

the con-

tents

of

the

proposed

letter

to

someone,

you

would

find

any

number

of

French

people

who

would

write

this letter

for

you

in correct

French.

Or else, you

would have

to

find

a

French

person

who

knew

enough

Enghsh

to

understand

your

purpose.

P.

U.

—

Don't you

think that

this

lack

of

under-

standing,

this

chasm

between

theory

and

practice

in

this

country,

is

gradually

closing

up?



A

DIALOOUB.

xm

A,

—

I

most

certainly

do,

and

I will tell you the

why.

At

first

we

had

in

this

country

only

^Ictical

practicians

and theoretical

theoreticians.

They were

like

two persons

who

spoke

different lan-

guages

and did

not

care

for

each

other's

point

of

view.

Now

we

have

two' classes

of

middlemen in the

field:

theoretical

practicians

and practical theoreticians.

This

applies

not only

to engineering

and

mathematics,

but to

nearly

all

forms of

human

activity.

A

practical

theoretician

is

usually

a

professor

of

applied

mathe-

matics or

physics,

who

fully

understands

the

langua^

of

the

theoretical

mathematician

and

at the

same

time

is

full

of desire of

applying the

results

of the theor>'

to

the

solution

of practical

problems, though

he

may

not

know

exactly

what

is

needed

in

practice.

A

theoretical

practician is

usually

a

professor

of

engineering,

or an

expert

in the

employ

of some

industrial

cori^ration.

He is

fully

conversant

with the

practical

conditions,

and

sees

the

need of

solving certain

practical

problems

in

a

rational

way

by

means

of

mathematics, but he

is not

enough

of a

mathematician

to tackle

these

prob-

lems

himself.

Now,

the

theoretical

practician

and

the

practical

theoretician

have

enough

in

conmioii

to

understand

each

other,

to

sympathize

with

each

other,

and to

solve

problems

cooperatively.

In so doing

they draw

for

information

upon

the

theoretical theo-

retician

and

the

practical

practician,

and

thus

bring

these

two

classes

of

pef)plo

to

ser\'e

each

otheri

P. U.

—

Then,

using

your

analogy,

I

would say

that the

theoretical

theoretician

is

a

man

who speaks

French

only,

and

a

practical

practician

is a

man

who

speaks

English

only.

.\

practical

theoretician

is

a

Frenchman

who

is fond

of English

and

dabbles

a

little

in

it.

A

theoretical

practician

is

an

Englishman

who



XIV

A

DIALOGUE.

loves

French

and understands it

a

little

when

it

is

not

spoken

too

fast.

These

two men

become

friends,

and

through

them

the French-only

person

is

made

to

understand

the '*

nothing-but-English

individual.

A.

—

This is exactly

what I

meant.

P.

U.

—

Well, now

that I understand the

purpose of

your

book

and

the

method

of

studying it, the task

does

not seem

so

difficult

to

me.

I

guess

I'll

get busy

at

once

on

that

first

chapter

about

the

incHned

plane.

A.

—

Yes, and

see

to it

that

you

climb up

that

plane,

and not

roll down.



CONTENTS.

Paob

I.

Incunkd

Plane

and

Screw

1

II.

Friction

IN

JouENAU

U

III.

Friction

in

Step

Bearinos

19

IV.

Carrtino

Capacitt or

Beltb

32

V.

Torsion

of

Shattb

40

VI.

Moment

or

Inertia or

Fltwheels

48

List

or

RatmREMCE

Works

64

Appendix

ti7

KT



[GINEEKIN

(

J

M

A IH

1

:^[

ATICS.

CHAPTER I.

INCLINED

PLANE

AND

SCREW.

LOAD

L,

weighing

P

kg. (Fig.

1),

is

hauled

up an

mclined

plane

AB

&t

&

uniform

speed,

by a

horizontal

force

of

Q

kg.

It

is required to find

0,

the angle

of

inclination

of the

plane,

such

that the mechanical

effi-

ciency

of the

arrangement may be a

maximum.

The

mechanical

efficiency

is

defined

as the ratio of

the

useful

work

jjerformed to the

total

energy

expended.

Let the

load move

a

distance

8

along

the inclined

plane;

the

useful

work

performed against

the force of

gravity P is

then

P

•

s

sin

6,

where «

sin

6

is

the distance

traversed

by

the

load

L

against

the direction

of

the

resisting

force

P.

Similariy,

the

work

expended

by

the moving

force

Q

is

Q'scosd.

According

to

the

definition

given

above, the efficiency

Ps sin ^

P.

. /,v

^

=

-r

=

y.

tan

(1)

The

amount

of

work expended by

the moving force

Q

is larger

than

that

merely required to

Uft

the

weight,

because

the

force

Q

has,

in

addition, to overcome

the

friction

between the load and the

plane.

In order

to

determine

the

ratio between P and

Q,

it

is

therefore

necessary

to consider the

force of friction. This

force

is found, by

experiment,

to be proportional to

the

nor-

mal

pressure

between

the

rubbing

surfaces,

and

to

be

1



ENGINEERING

MATHEMATICS.

[Chap.

I.

directed

so

as

to

oppose the motion.

In

Fig.

1,

N

represents

the

normal

pressure

of

the

plane

upon

the

load,

and

nN

is

the

corresponding

force

of friction.

The factor

of proportionahty

/i

is

called

the

coefficient

of

friction; its

numerical value

depends upon

the

mate-

rials, finish,

and lubrication of

the rubbing

surfaces.

Diagram

of forces for

an

inclined plane;

the

load

being

moved

upward by

a

horizontal

force.

The force

R,

being the

resultant of

N

and

nN,

repre-

sents the

total

reaction

of the inclined plane

upon

the

load. It will

be

seen

that

this

resultant

is

inclined at

an

angle

from the

normal,

such

that

-N=^

an

<t>

(2)

This angle is called the

angle

of friction

;

the amount of

friction

between

two surfaces can

be

indicated by

giving

either

the value of /x or the

angle

^,

the

two being

connected

by

Eq.

(2).

The load P is

moving,

by

as-

sumption, at

a

uniform rate under

the

influence

of

the forces

P,

Q,

and

R;

hence these three

forces

must

be

in

equilibrium.

In

other

words,

the

resultant,

LTy



Chap.

L)

ISCUSBD PLANE

AND

SCREW.

Z

of

P

and

Q

must be

equal

and

opposite

to

the

force

R.

It

will

be

seen

from

the

hgure

that

the

angle

PLT

«

+

^,

so

that

<?-Ptan(tf4-*)

(3)

Substituting

this value

of

Q

into

Eq.

(1)

gives

tan(«

+

0)

Thus,

the efficiency

depends

only

upon

the

angle

of

inclination

$

and

upon

the

angle of

friction

0,

and

is

independent

of

the

weight

of the

load. To

find

the

value of

6,

for which

the

efficiency

becomes

a

maximum,

we

equate

to

zero the

first

derivative

of r,

with

respect

to

d.

This

gives the

following

equation

for

6:

^

tan

(«

+

^)

=

_,

J

.

—

r

.

tan

d.

.

(4a)

cos*

e

C08«

{0

+

0)

/ The

efficiency

ij

is

a

maximum when

d

satisfies

/

this

equation (see

also

Prob.

22

below).

Replacing

]

the

tangents

by

the

ratios

of the

corresponding sines

(

and cosines,

we

find

after

reduction

that

sin2(«+0)

=

sin 2«.

Now

the

sines of

two

angles

are

equal,

either

when

the

ang^

themselves

are equal,

or when the

sum

of

the

an^es

is

180

degrees.

The

first assumption

leads

to

^

—

0,

and

thus

m

0;

in

other

words, there is

no

friction.

In

this

case

the

efficiency

is

equal

to

100

per

cent

at

any

value of

d, since

there

is

no

source of

loss

of

work.

Moreover,

in

this

case

n

in

Eq.

(4)

is

a

constant

->

1,

so

that

there

is

no

question of a maxi-

mum.

The

second

assumption

gives

2

(tfi

+

0)

+

2

9i

-

180

degrees,

from

which

«.-45*»-|

(5)

Herv

di

is

the

value of

d

for

which

r^

is

a

maximum.



4

ENGINEERING

MATHEMATICS.

[Chap. I.

The smaller

the

friction

the nearer

di

approaches the

value of

45 degrees;

the presence

of friction reduces

the

value

of

this

angle

of

inclination. Thus,

for in-

stance,

if

the angle

of friction

</>

=

10

degrees,

the best

angle

to

use

for

an

inclined

plane is 40 degrees;

if

^

=

20

degrees,

di

=

35

degrees, etc.*

/

Prob.

I.

—

The

coefficient of friction

between

cast

iron and

steel,

under

certain conditions,

is

0.347.

At

what angle

Oi

is the

efficiency

of

the inclined plane

a

maximum? Ans.

35°

26'.

Prob. 2.

—

Show that

the

Eq.

(5)

for

the

angle of incUnation

«i

is

equivalent

to

the relation

cot

2

0i

=

/i)

check the

answer

of

Prob.

1

using

this formula.

V

Prob.

3.

—

Show that

T/max

=

tan''

9^,

by

substituting

Eq.

(5)

into formula

(4).

Prob.

4.

—

Plot curves of

the most advantageous

angles

of in-

clination,

and

of

the

corresponding

efficiencies,

to

values

of

coefficient

of friction

as abscissae,

between

the

hmits

of

At

=

0,

and

n

—

0.5.

Ans.

The

extreme ordinates

of

the

curves

are:



Chap.

I

INCUNBD PLAXB

AND

SCREW.

uniform

speed. The

force

of

friction

mA'

(Fig.

I)

is

now

directed

^Opwmrd,

instead of downward,

and the angle

PLT

«

«

-

^. The

P

is

in tlus

case

the

moving

force,

and the efficiency

.

^«»*

.

Q

.

_L_

.

tan

(»

-

)

Ptsin*

P

tan#

tan#

(4aJ

Pi(?. J.

-

< ir:iphiraJ

determination of efficiency, ij,

armniing

to

Eq.

(4).

Find the mayimnm value of

this efficiency

at

a

given

angle

of

friction

and

with

various inclinations

of

the plane

to

the horizontal.

Atu.

f'n^

-

cot*»,,

where

»i

-

45°

+^.

Prob.

9.

—

A

load

is moving

up

and

down an

inclinc<l

plane, or

is

being raised and

lowered

by means of

a power

screw.

Show

that

the

efficiency

is

higher

when

the

load

is

going

up,

provided

that the

angle

9

is

less than

45

degrees.

Solution.

—

It

is

required to prove

that

tan

«

.

tan

(«

-

^)

or

that

when tan

f

is <

1.

We

have

tan

(8

+

^)

tan

tan*

«

>

tan

(9

+

^)

tan

(«

-

^),

1

-

tan'

#

tan*

4

^

1

-

tan*

/tan*

1

-

tan*

tan*

$

tan>9.

Since tan

•

is

<

1,

the quantity

subtracted

from

the unity

in

the

numerator

of

the fraction

is

larger

than

the

quantity

suhtrutted

from

the unity

in

the

denominator.

Hence

the fraction

in

]no]x'r,



6

ENGINEERING

MATHEMATICS.

[Chap.

I.

and

the inequality holds true. In

a

similar

manner,

when

e is

above

45

degrees,

tan »

is

>

1,

and the fraction

is improper.

The sign

of

the

inequality

must

then be

reversed;

this

means

that

in this

case

the

efficiency

is

higher when

the

load

is

descending.

Prob.

10.

—

A

load

is

being

hauled

up

an

inclined

plane

by

a

force

parallel

to

the

plane

(Fig.

3).

Find

the inclination

of the

plane

at

which

the

efficiency

reaches

its

maximum.

Pig.

3.

—

Diagram of forces for an

inclined

plane;

the

load

being

moved

upward

by

a

force

parallel

to the

plane.

Ans.

The efficiency

reaches

its

maximum

value of 100

per cent

when

the plane is vertical;

in this

case

there is

practically no

pressure

and

consequently no friction between the

load and

the plane.

Solution.

—

When

the load

has traveled

a

distance

s

along

the

plane,

the

work

done

against

the

force

of

gravity

is

P

•

s

sin

e,

while the

work

expended by the

moving

force

Q

is

Qs.

Therefore

the

efficiency

Ps sin

d

P sin fl

V

=

Qs

Q

But from the

triangle LPT

so

that

P

Q

sin

(90°

-

<^)

sin

(^

+

<^)

V

=

1os

(^

sm e

sin {e

+

<t>)

1

+

tan

«/»

cot e

(6)

The

efficiency is

the

nearer

100 per cent, the smaller

the

product of

tan

<t>

'

cot

e.

This

product is equal

to

zero

when

either

<^

=

or

e

=

90°.

The

first

case corresponds

to an

ideal plane without

friction;

the

second

assumption

gives

a

vertical

plane.



Chap.

I]

ISCLISBD

PLANS

AND

8CRKW.

Prob.

II.

—

Referring

to

the

preceding

problem,

let the kMd

move

down

the

plane.

By

analogy with

Prob.

8,

the eflkiency

is

V-?BjLz^.l^tan4oot«

coe^sm

9

(6a)

Prove

that

i;

>

n',

that

is

to say, the efficiency

is

higher

in

ascend-

ing

the

plane than it is

in

descending.

Hint :

1 >

1

-

A'<;

conse-

quently

1

>

1

-

.V.

Another way to

prove the

proposition

(1+JV)

is

to

eaqpand

tf

into

an

infinite

series

by

dividing

1

by

(1

+

iV).

It

must

be

remembered that the

exprenion for

ij'

presupposes

that

Fig.

4.

—

Diagram

of

forceti for

an

inclined

plane;

the

load

being

moved

upward

by

a

force

inclined at an angle

a

from

the

horizontal.

tan

•

cot

«

<

1

;

otherwise

the definition

of efficiency

given

above

would

not

hold without

an

additional

qualificatiou.

Prob.

I

a.

—

A load

moves along

an

inclined

plane at

a

uniform

iqpeed;

the force

Q

(Fig.

4)

acts

at

an

angle

a from

the

horizontal.

Fmd

an

cj^reasion

for the

efficiency

of

the

mechanism,

when

the

load

moves up

the

plane;

and

also

for

the

case,

when

the

load

is

moving

down the

plane.

SohdUm.

—

When the

load

moves up the plane

(Fig.

4),

P resists

the

motion

and

Q

is the

moving force.

The

efficiency

is

then,

by

definition,

the

ratio

oi

the

useful work

performed

against

the

force

/'

to

the

woric

actually

expended

by

the moving

force

Q;

or

P<»coe(P,«)

Psin»

'

Qs.

cos

(g,«) gcos

(•-«)•



8

ENGINEERING

MATHEMATICS.

[Chap.

I.

But from the

triangle

of forces LPT

P

^

sin

LTP

^

cos

(e

+

<t>-a)

Q

sin

PIT

sin

(6

+

^)

so that

=

cos

(e

+

<t>

—

a) sin ._v

sin

(«

+

0)

cos

((9

-

a)

When the force

Q

is

horizontal,

a

=

0,

and the foregoing expres-

sion

for efficiency

becomes identical

with

Eq.

(4).

When

Q

is

parallel

to

the

plane,

a

=

d, and

Eq.

(7)

checks

with

Eq.(6).

When the

load

slides down the plane,

by analogy

with

Prob.

8

we

get for the efficiency

/

^

sin(g-<^)cos(g-tt)

^

V

'

cos{e—<t>—a)s\ne

This

expression

reduces

to Eq.

(4a)

when

a

=

0,

and to

Eq. (6a)

when

a=

e.

Prob.

13.

—

For the values

of

di

and

<f>

found

in

Prob.

1,

plot

an

efficiency curve, when

the angle a,

at

which the force

Q

is

acting,

varies from

+

90

degrees

to

a

negative value

at

which

the

efficiency

becomes

zero.

Ans.

a

=

+

90°

+40°

0°

- 35°

26'

77=

100 69.1

50.6

Prob.

14.

—

Determine

the

value

of

the angle

of

inclination

e

in

Eq.

(7),

at

which

the efficiency becomes a

maximum, with

given

values

of

and

a.

^ns.ei

=

45°

+

|

-|.

Hint.

—

Use

the transformation

2

cos

AsinB

=

sin (A

+

B)

—

sin

(A

—

B),

before

differentiating

Eq.

(7).

Prob.

15.

—

Using

the

answer

of

the

preceding

problem,

prove

that

_

sin^

fli

cos''

(^1

—

a)

and

show

that the

result

checks

with

that

obtained

in

Prob.

3,

when a

=

0.

Prob.

16.

—

Assuming

m

=

0.25,

plot

a

curve

of

values of

^1,

the

most

advantageous

angle

of

inclination

of

the

plane,

for

various



Chap. I

)

L\CLL\ED

PLASB

AND

SCREW,

9

ftn^tos

a

of hauling: aLto

a

curve

of

the

corresponding

values

of

eflleiency.

Ans.

a

-

-70

+76*

+90*

»

-

38

76

83

i?m*x

-

61

94.1

100

Prob.

17.

—

At

what

angle

a

must

the

force

Q

be applied

in order

to

obtain

a

maximum efficiency

on a

given

inclined

plane?

Ant. The

efficiency is

100 per cent when

the

load

b

simply

lifted

vertically;

when

Q

acts at

an

angle

to

the

vertical,

the

efficiency

b

the

higher,

the

near^

the direction

of

Q

approaches

to

the

vertical.

Solution.

—

The

efficiency

b a

maximum

when

the term

000 (#

+

4

—

a)

•

«-.

.m\

•

i-.'-^

-~-m

Eq.

(7)

becomes

a

maxmiium.

cos

\9

—

a)

Hut

cos

(»

+

—

a)

* /. \

,;. ,

« COS

-

sm

#

tan

(»

-

a),

cos

IS

—

ai

SO

that

f

b a maximum when tan

{9

-

a)

has the

largest

possible

negative value consistent

with

the

problem.

Thb c<NTesponds to

a

-90*.

Prob.

18.

—

The preceding theory

applies

essentially

to a

square-

threaded

.screw, which

can

Ix?

con«iidered

as an

inclined

plane

cut on

a

cylinder.

However,

in

calculating

the

^ciency

of

a

screw,

it is

necessary

in

some

cases

to

take

into

account

the

friction

in

the

pivot

or collar, in

additbn

to the

friction

lietween

the

screw proper

and

the

nut.

Wlien the

driving force

Q

Ls

horizontal, this

additional

ooUar friction

b

proportional

to

the

weight lifted,

and

can

be

repre-

sented by

a

horixontal

force//*,

where/ is a constant depending upon

the

coefficient

of

pivot friction

and upon the

dimensions

ci

the

pivot.

The

problem

is

to

fitid an

expression for

efficiency,

analogous

to

Eq.

(4),

when the

screw is lifting

a

weight

/'.

Am.

Eq.

(3)

becomes:

Q-

PUn(«

+#)+/P,

so that

tan#

rgv

''-t-iMrMTT/

^^^

Prob.

19.

—

Find the

value

of

the

thread

angle 9 which converts

the expresrioa

(8)

into

a

maximum.

Ant.

tan»,--it+vA*+*.'^^reM-tan*andit-;^?^^r.

V

M

(1

-M/)



10

ENGINEERING

MATHEMATICS.

[Chap.

I.

Hird.

—

Denote

tan

6

=

z,

so that

'

+

'^+/

1

—

2/i

introduce k into

this

expression and

differentiate

with

respect

to

z.

Prob.

20.

—

The

coefficient

of

friction

m

between

the screw and

the

nut

is

estimated

to

be

0.18 for

a

certain

kind

of

service.

Tabu-

late

values

of

the

best

tliread

angles

and of the corresponding

efficiencies,

for values

of

/

from

to

0.20.

Ans.

/

=

01

=

39°

54'

77

=

69.9

0.20

48°

40'

61.3

Prob.

21.

—

When the movement

of

the screw

is

downward we

find,

by

analogy

with

Probs.

8

and

18,

_.,_

tan(g-0)-/

•n

=

-—

•

•

tan

d

Find

the

value

of

B

at

which

the

efficiency

reaches its

maximum.

Ans.

tan

»2

=

A;

+

d

k^

+

-

(see

Prob.

19).

Prob. 22.

—

Eq,

(4a)

represents

the

numerator

of

expression

dfi/dd

equated

to

zero. But

drj/d9

is

also equal

to

zero

when

its

denominator is

equal

to

infinity, that is,

when

tan^

{9+

4>)

=

QO

Show

that this value

of

e does

not

correspond

to

max. i?.



CHAPTER

II.

FRICTION

IN

JOURNALS.

A

iioHizoNTAL shaft

(Fig.

5)

revolves

in

a bearing

that

fits

it

well.

The problem is to determine the force

of friction

between

the

bearing

and the journal (the

part

of a

shaft

inside

of

a

bearing

is

usually

called

the

journal).

This

force

of

friction,

which

is tangential

to

the

surface

of

the

journal and oppxises

the

motion

of

the

shaft,

is usually assumed

to

be

proportional

to

the

normal

pressure between

the bodies

in

contact.

In

the

ca'^e

under

consideration,

this

pressure

may

be

diflFerent in

diffenMit

places

on

the

journal,

so

that

the

total

force

of friction is

obtained by

integrating

the

infinitesimal forces of friction

over the total

surface

of

the

bearing.

Let the

normal

pressure per

square

centimeter

of

the

surface

of

the

bearing be

denoted

by

p.

This

unit

pressure,

generally

speaking,

is

different

for

different

horizontal

strips

of

the

bearing,

the

weight

of

the

shaft

being

distributed

non-uniformly.

In

mathematical

lan-

guage,

p

may

be

said

to

Ix*

a

function

of the

angle

B

(Fig.

5).

The

normal

force

acting

upon an

infinitesi-

mal

strip,

/

rde, of

the

bearing

is

plrdd;

the

correspond-

ing

force

of

friction

in

up

I

rde,

where

m

is

the coefficient

of

friction.

The

numerical

value

of

n

varies

within

wide

limits

according

to

the

materials

used,

and

to

the

kind

of lubrication.

Integrating,

gives the

total

force

of

friction,

F''2

I

uplrdd'A

f\pde.

.

.

.

(1)

11



12

ENGINEERING

MATHEMATICS. [Chap.

II.

Here

A

=

2rl

is

the

so-called

projected

area

of

the

bearing,

or

the

cross

section of

the

journal

by

the

horizontal

plane

passing

through

its

axis.

The

use

of

Fig.

5.

—

The

journal of

a

shaft

resting

in a

bearing.

this

projected area

is

convenient in

practical

calcula-

tions

pertaining to

journals

and bearings.

The

limits

of

integration should

be

+

^

and

—

^

, but since

the

bearing

is

symmetrical

it

is

sufficient to

calculate

the

force

of friction

for

one

quadrant

and multiply

the

result

by

2.

Assuming

the

coefficient of

friction

to

be

independent

of

the

pressure

p,

Eq.

(1)

is simplified

to

F

IT

=

Afi

I

p

dd.

(2)

In order

to

perform the integration,

p

must

be given

as

a

function

of

the angle

6.

The

actual

distribution

of the

pressure

over

the

surface

of the

bearing

is

not

definitely

known. Various

reasonable

assumptions

in



n.l

FRICTION

IN

JOURNALS.

1

•

regard to

this

distribution

are made in

the

problems

that

follow.

But

whatever the

distribution of

pressures, the

sum

|total

of

the

vertical

projections of these pressures must

^be

equal

to the

load

P

resting

upon the bearing,

in

porder

that

the

bearing may

be in

equilibrium.

This

gives a

general condition

which the

unit

pressures must

satisfy

in all

cases;

namely:

w

w

P^2

f'plrdeco6d''A

l*pcoaede.

.

.

(3)

All the

problems that

follow

are

solved

by the use

of Exj.

(2)

and

(3),

when

the

coefficient

of

friction

is

assumed

constant,

or

by Eq.

(1)

and

(3),

when this

coefficient

is

variable.

Prob.

I.

—

In a new

bearing

with

a perfect

fit the

normal pressure

p

can

be

assumed

the

same over

all the

surface

of

the journal

;

in

other

words,

p

«

Pf

•

Constant (4)

Express the f(^cc

of friction

f

as a

function

of the

load

P

and

the

coefficient of

friction,

m-

Solution,

—

Eq.

(2)

gives

F-.4mP.-^,

(5)

and from

Eq.

(3)

P-.4p,

(6)

Combining

these two

equations

in

order

to

eliminate

the

unknown

pressure

p»,

we obtain

F-mP-^

(7)

Prom this formula the

resisting

fcMrce

of friction

caujMxi

by

a

hearing

can

be calculated,

when

the load P

resting

upon

the

bearing,

and

the

coefficient

of

friction,

n,

arc

known.

Prob.

a.

—

Boarinflps

are

always

made

of a

softer

metal than

shafts,

so that,

in

proportion

as

the

bearing

wears

out,

more

and

w.jrt

pressure

is

exerted

on the bottom

of

it,

and lecw on

the Hides.

Afto*

the bearing

has

been

in use for

some

time,

it

is

ground

by

the



14

ENGINEERING

MATHEMATICS.

[Chap. II.

journal

into

such

a shape

that

further

wear

does

not alter

its

shape,

but

merely lowers

the

position

of

the shaft.

Let

S

in

Fig.

6

be

this

final shape

of the

surface

of the journal.

The

surface

S'

to

which

the

bearing

is

worn

after

some

further

use

is

identical

with

S,

the

Fig.

6.

—

Two

consecutive

surfaces of

a

bearing, with

a constant

vertical

wear.

vertical

wear w

being the

same at all

points. The

problem

is

to

find

the

distribution

of

the

normal

pressures

p

such

that gives

a

uniform

vertical

wear.

Solution.

—

Let

the normal wear

at

a point

K,

defined

by

the

central

angle

d,

be

n.

This

wear

can

be assumed

proportional

to

the normal

pressure

exerted

at

this

point;

*

in other

words,

n

=

Gp.

Here

G

is

a constant

which depends

upon the

material

of

the bear-

ing,

the

lubrication,

and the speed of

the

shaft.

On

the

other

hand,

we have

from

the

figure

: n

=

w cos e, so

that

w

cos 9

=

Gp. Since,

by

assumption,

both

G

and

w are

constant, we have

p

=

Po

cos

e,

(8)

where

po

=

—

is

a

constant.

This is the required

law of

distribu-

9

tion of

the normal

pressure

in

a

worn

bearing.

When e

=

0,

p

=

po,

so

that

the constant

po

represents the unit pressure

at the

bottom

of the

bearing. The pressure

p

decreases from

the bottom

of

the

*

Normal

wear is

proportional

to the work

of friction

per

unit area,

so that at

a given

speed,

and

with

a constant

coefficient

of friction, it

is

proportional

to

the

normal

pressure.



r.

11

FRICTION

IN

JOURNALS.

16

bearing

upward

aa

the

ooone

of the angle

9.

No normal

preaaure

m

exerted

at the

aides,

where

9

«

90*.

Prob.

3.

—

Find

the relation

between the

load

and the

force of

friction

in

a

worn

bearing,

using

Eq.

(8).

Solution.

Substituting

the value of

p

into

Eq.

(2)

and

(3)

and

we

find

F

-

-4mPo

(9)

P'Ap,-'^

10)

Eliminating

Ap% gives

F'^P'-

(11)

Prob.

4.

—

A

journal

15 cm.

in

diameter and

32 cm. long

runs at

a

speed

of

200

rev.

per

min.

The

weight supported

by the

bearing

is

12

tons (one

metric

ton

>

1(XX)

kg.).

Assuming the

bearing to

be worn, calculate:

(a)

the power lost in friction, with

a

coefficient

of

friction

«•

0.0032;

(6)

the

average

pressure

upon

the

projected

area;

(c)

the maximum unit pressure

at

the

bottom

of the

bearing.

Solution.

—

(a)

Using

formula

(11)

we

get:

f-

0.0032

X

12,000

X

-

-

48.89

kg. The path traversed

by the force

of

friction

per

aecond is

—

—

^-

1.571

m., so

Umt

the

work

lost

is

48.89

60

X

1-571

«

76.8 kg-m. per

sec.

« 753

watts

(one

kilogratn-meter

per

P

12000

aecond

is

equivalent

to 9.81

watts).

(6)

—

«

77=-^^;^;

25 kg.

A

(15

X

SIS)

4

per

sq.

cm.

(c) From Eq.

(10),

p.

-

25

X

-

-

31.8

kg.

per sq.

cm.

V

Prob.

5.

—

Solve Prob.

4

under the supposition that the

bearing

»

new.

Anj>.

(a)

930;

(6) 25;

(c)

25.

Prob.

6.

—

Solve Prob. 3

with

the assumption

that

the pressure

varies according

to

the

law

p

-

p«

cos*

9.

Ans.

F-mP-^;

P'Iap..

Prob.

7.

-

Calculate

friction in

a

bearing

in

which

normal

pressure varies

aocwding

to

the straight-line

law

p

p«

(

1

-

h9),

where

Jk

is

an empirical

eoostant.

An*. P-^p.(l-*^

+

l-V

'-''[^-*FVC'-*a-')]-



16

ENGINEERING MATHEMATICS.

[Chap.

II.

Prob.

8.

—

The

force of friction

in

a

bearing

was

found

from

experiments

to

be

expressed by

the relation

F

=

1.35

nP.

Assum-

ing

the

pressure

to

vary

according

to

the

straight-Une

law, as

in

Prob.

7,

determine

the

value

of

the constant

k,

and

the

ratio of the

smallest

to

the greatest unit

pressure.

Am.

k

=

0.478;

^5»B

=

i

_

A;

^

=

about 0.25.

Po

2

Prob.

9.

—

Solve

Prob.

7,

assmning that

the pressure varies

as

the

square

of the angle, that is

to

say,

according

to

the formula

p

=

Po

(1

—

ce'^),

where

c

is

an

empirical constant.

Ans.

P

=

^po[l-cf^-2^];

Prob.

10.

—

A

bearing

having

a

projected

area

of 300

sq.

cm.

supports

a

weight of 6 tons.

Calculate the

maximum pressure at

the

bottom

of

the bearing

under

the

following

assumptions:

(a)

The

bearing is

new;

(&)

the

bearing

is

perfectly

worn; (c) the

pressure

varies

as

cos*

d; (d) the

pressure varies

according

to

a

straight-line

law,

the

pressure

being

zero

at

d

=

90°;

(e)

same

as

(rf),

only

the pressure at ^

=

90°,

is

equal

to

one-half of the

maximum

pressure;

(/)

the pressure varies according to a

parabohc

law, as

in

Prob.

9,

the

pressure

being

zero

at

=

90°;

{g)

same as

(/),

only the

pressure

at

=

90°

is

equal to

one-half of

the

maximum pressure.

Ans.

(a)

20;

(6)

25.45;

(c)

30;

(d)

31.42;

(e)

24.45;

(/)

24.67;

(gf)

22.1

(in

kg. per

sq.

cm.).

Prob. II.

—

Using

the

data

of

the

preceding problem, plot

curves

of

the distribution

of

pressure

against angles d as

abscissae.

Draw

all

the seven

curves

on the

same

sheet

and

to

the

same

scale,

to

enable a

direct comparison.

Ans.

At »

=

45°,

the

ordinates

are:

(a)

20;

(b)

18;

(c)

15;

(rf)

15.71;

(e)

18.34;

(/)

18.50;

(g)

19.34

(in kg.

per

sq.

cm.).

Prob.

12.

—

Show

that

the

curve

p

=

po

(1

—

cd^),

used in

Prob.

r,

10,

and

11,

is

a

parabola;

reduce

its

equation

to the

standard form w*

=

2

mx.

Ans.

0 ^

=

2'

—

•

(po

—

p).

2poC

Prob.

13.

—

In all the foregoing

problems the

coefficient

of

friction

is assumed

to be

independent

of the

pressure. Some

experi-

ments

indicate that

withih

certain

Umits

the

coefficient

of

friction

decreases

with

increasing

friction.

Solve

Prob.

3

under

the

assump-



C»Ay.

11.)

FRICTIOS

IS

JOURNALS.

17

tion

that

the ooefficicDt of frictioii

varies

aeeording to the

kw

0

0*-

/p,

where

m

»nd

/

are

empirieal

oonataats.

Ana.

F

-i4p«fM«-p«<'T].

where

Pais

determined

by

Eq.

(10).

HmL—Vm

Eq.

(1)

instead

of

(2).

Prob.

14.

—

Solve

Prob.

13

under

the

suppodtion

that

the

co-

efficient

of friction

m

is

directly

proportional

to

the square

root

of

the

peripheral

qieed of

the

journal and

inversely

proportional

to

the

square

root of the unit

|Heasure.

Aiu.

tfm

i/-

,

where

mi

ia

A ooosUnt;

>

P

F-^Mi'Vi^,

I

Vooe

i.(i9-

1.19

A^v^

The

maxJmu

ni pressure

p«

is

determined

from

Eq.

(10).

NoU.

—

VcoB0

cannot

be

integrated in

the

finite

form, and

must

be

calculated approximately, for example by

means

of the

trape-

soid

formula,

or

by

the Simpson

rule

(see

also Prob.

17).

Accord-

ing

to

the

former, we

have,

taking

ordinates every

5

degrees:

X'

v^«#.d#

--S^X

j~Q

VcoaO^

+

Vw^

+

V008ICP+

• •

+

VoosSS*

+

^

VcosWrJ

-

1

Prob.

15.

—

C alculate

the

integral

in

the

preceding

prr>blom,

using Simpson's rule, to see how

closely

the result

checks

with

that

obtained

by

the

use of

the

trapezoid

fcmnula.

Evaluate

the

same

btegral

by the two

methods,

taking

larger intervab

—

for instance,

10,

and

15

degrees.

This

will

show

the

advantage

d

%npson's

*

The

solution

is also obtained

directly

through

the

Gamma

func-

tion.

According

to

formula

483

on

page

92

of

Peiroe's

table

of integrals,

we

have

r

or,

nnoe T

(n

+ 1)

-

nT

(a),

Valnss

of

the

comroonloganthms

of

r(ii)

for

values

da

between

1

and

2

are tabulated on

page

124

of Psiroe's

book.



18

ENGINEERING

MATHEMATICS.

(Chap.

11.

*

rule

over the

straight-line formula,

in

that

more

accurate

results

are

obtained with a

smaller

number

of

points.

Prob.

i6.

—

In the solution of

Prob.

14 and

15

it is

assumed

that

the

relation

(8)

holds for perfectly

worn

bearings,

when

the

coefficient

of

friction

is

variable. Strictly

speaking,

this is not

true,

because

Eq.

(8)

is deduced

under the

assumption of

a

constant

coefficient

of

friction. Show that,

when

the

coefficient

of

friction

vaxies

as

in

Prob.

14,

the

distribution

of pressures

in

a perfectly

worn

bearing

is that

assumed

in Prob.

6.

Solution.

—

The

normal wear

at

a

point

is proportional

to the

work

of

friction per

imit area. Since

the velocity

is

the

same at all

points of

the

journal, the wear

is

proportional

to the

friction

force

per

unit

area.

This force of

friction is in our

case

np

=

m

'^vp,

so

that

analogously

with Prob.

3

we have

n

=

w cos

e

=

Cm

Vvp,

where

C

is

a

constant.

When

d

=0

the

foregoing

equation

gives

•

w

=

Cm

^vpo.

Dividing

one

by

the

other to

eliminate

C

and

w,

gives

cos

»

=

v/

—

or

P

=

Po

cos2 0.

Prob.

17.

—

T

he

definite integral in Prob.

14

can also

be

evaluated

by

expanding

Vcosfl into

an infinite

series

according

to

Maclaurin's

theorem

and

integrating

the series

term

by term. Compute

the

first

few

terms

of

this

expansion and

show

that the series is rapidly

converging. Discuss

the advantages of

this

method

as compared

with those

shown

on

the

preceding page.



CHAPTER

III.

k

FRICTION

m

STEP

BEARINGS.

A

VERTICAL

shaft

M

revolves

on

a

step

bearing

iV

(Fig.

7)

;

it is

required to find

the work

lost

in

friction

between

the

shaft

and

the bearing

per

one

revolution

of

the

shaft.

This

problem

is similar to

that

of

friction

in

horizontal journals,

considered

in

Chapter

II;

only

here

it

is necessary

to deter-

mine the

vxyrk

of friction,

w^hile

with horizontal

jour-

nals

it

is

sufficient to eval-

uate

the

force

of

friction.

This difference

is

due

to the

fact

that

in a

bearing sup-

porting

a

horizontal

shaft

(Fig.

5)

all

the

elementary

forces

of friction are

at

the

same

distance

from

the cen-

ter

of

the

shaft, and

hence

their

points

of

application

move

^^ith

the

same veloc-

ity;

while

in

a

step

bearing

the

points

of

application

of

dementary

forces

of

friction

are

moving at

different

velocities,

from

zero

on

the

center

line

of

the

vertical shaft,

to

a maximum

on its

peripher>'.

Therefore

the forces cannot

in

this

case

be

simply

added

together,

but

the

share

of

each

ele-

fOdnXajy

force

of

friction

must

be

determined

in

the

10

Fig.

7.

—

A

flat

step

bMring

mip-

pofting

a

vertical

shaft.



20

ENGINEERING

MATHEMATICS.

[Chap.

III.

proportion in which it

contributes to

the

total

resisting

moment,

or

loss

of

energy.

The

student

is

supposed

to be

familiar

with

the

prin-

cipal

problems

solved

in Chapter

II;

therefore,

some

explanations are

omitted

to avoid

repetition.

Prob.

I.

—

Deduce for a

flat

step

bearing

(Fig.

7)

expressions

for

the work

of

friction,

and

pressure,

analogous

to Eq.

(1), (2),

and

(3)

of

Chapter

II.

•

Solution.

—

Let

the

pressure

per

square

centimeter

at

a

distance

X

from the center of the shaft be

p;

the

total force

of friction

upon

the

infinitesimal

zone

2irX'dx

is

then

np-2irx-dx,

where

n

is

the

coefficient

of

friction. The

path traversed

by

this

force

during

one

revolution of the shaft

is

2 nx, so that the work

of

friction

per

revo-

lution, upon

the

infinitesimal zone,

is

up

'

2 irX

'

dx

'

2 irX

=4

ir-fipx^

dx.

Total work

of

friction per

revolution

W

=

iTr^

r

fxpx^dx

(1)

If the

coefficient of

friction

can

be

assumed

to

be

independent

of

the velocity

or pressure, the

foregoing

expression

is

simplified

to

W

=

^Tr^f,

I

px^dx

(2)

Jo

The

unit

pressures

p

must

satisfy the

condition

that

their

sum

over the

entire

bearing

surface

be

equal

and

opposite

to

the

total

weight

P

resting

upon

the step

bearing;

or

2-irx-dx

(3)

As

in the preceding

chapter,

these

equations

can be

integrated

only when the

law

of

distribution of

the

pressures

p

is

known

or

assumed.

Some

specific cases

are considered in

the

problems

that

follow.

Prob. 2.

—

Find an

expression for the work

of friction in

a

flat

step

bearing,

assuming

the

bearing

to be

new,

that is

to say, the

pressures

are distributed

uniformly over the whole

supporting

area.

Solvtion.

—

The

given

condition

is

expressed

by

the

equation

p

=

Pav

=

Constant

(4)



CUAT. III.

FRICTION

IN

STBP BEARINGS,

21

Therefore

Eq.

(2)

becomen

W

•mPm

4rV

and

fram

Eq.

(3)

we find

^Eiiinitiaiitig

p«,

between

these

two

exprnwious

gives

4rr

W^uP'

3

'

(5)

which

is

the

required formula

for

the

work

of

friction

per

ooe revo*

lutioD

of the shaft.

.

This

formula can also

be

writteit

in

the

form

ir-iiP.2

(t)

which

can

be interpreted by

saying

that in

a

new bearing

total

friction

may

be

considered as

if

concentrated

at two-thirds

ci the

radius

of

the

shaft.

Prob.

3.

—

Fmd

the

law of distribution

of pressure in a

per-

fectly-worn

flat step

bearing;

make

use

of

the principle

explained

in

Prob.

2.

Chapter

II.

Solution.

—

The required

distribution

of pressures

is such as to

cause the

same vertical

wear at

all

points of the

bearing.

But the

wear is proportional

to

th<-

work vi

friction

per

unit

area

(experi-

mental

fact)

;

hence

Work

of

friction

_

4^^px»dr

_

Constant,

Area

2rxdx

at, after

reduction,

omitting the constant factors:

fipx

-

nrPrT

-

Cottatant,

(6)

where

^

and

p,

arc the

values of the coefficient

of

friction

and of

the

unit

pressure

at the periphcrj'

of

the

shaft. If

the

coefiicient

of

'

oniitant, the

foregoing

expi

'

easi

on

becomes

pz

PfT

^

Constant

(7)

Ib

iroRb:

the pressure in a

worn

step

lx>aring

varies invonely

as

tiht

^stance

from

the center of

the

diaft,

lieing

a

minimum

at

the

periphery.

Eq.

(7)

probably does

not

hold good for

points

near

the eeoter of the

shaft,

because,

when

z

approaches to lero,

p

ap-

proaehes

to

infinity.

There

is

do

doubt,

howevo-,

that

in

a

worn



22

ENGINEERING

MATHEMATICS.

[Chap. III.

step

bearing

the

pressure

near

the center

of

the

shaft

is

very

great.

In

practice,

the

shaft

always has a center

hole for the purpose of

centering

the

revolving

part

when

turning

it

on a

lathe,

so

that

there

is

no

pressure

at

the

center,

and

therefore no

objection

to

using

formula

(7).

Prob.

4.

—

Find

an expression

for the

work of

friction in a per-

fectly-worn

fiat step

bearing.

Solution.

—

Integrating

Eq.

(2)

and

(3)

with

the

use

of Eq.

(7),

we

find

IF=MPr-2

7rV

and

P

=

pr'2Tr'

(8)

Ehminating

pr

between

these

two equations

gives

W^fj^P'irr

(9)

Eq.

(8)

and

(9)

can also be

written in

the

form

The

following

conclusions

can

be

drawn from these

results,

as com-

pared

with the

results

of

Prob.

2:

(a)

The

total

work of friction

is

less in

a

worn

bearing

than in

a

new one,

in

the

ratio of

3

to

4;

(6)

the

resultant

force of

friction may be

considered

as

if

concentrated

at

one-half

of

the

radius

of

the

shaft;

(c)

the unit

pressure

pr

at

the

periphery

of the

shaft

is

equal

to

one-half of the

pressure

Pav

ob-

taining

in

a

new

bearing;

this means that the

parts of the worn

bearing near

the

center are

subjected

to

considerably

higher

pres-

sures

than

when

the

bearing

is

new.

Prob.

5.

—

The loss of

power

caused

by

friction in

a

step

bearing

can

be

expressed by

the

formula:

Px

Dx

N

Loss,

in

metric horse

power,

=

—=r

—

Constant

where

P

is

the load

upon the bearing

in metric tons,

D

is

the diame-

ter

of

the

shaft

in

meters,

and

N

the

speed

in rev.

per

min.

The

constant

is

placed

in

the

denominator,

in

order to make the formula

more

convenient

for slide-rule

computations.

Determine

the

value

of the constant

in

the

foregoing

formula for

a

worn

flat

step

bearing,

when

the

coeflBcient

of friction

m

=

0.03.

Ans.

Constant

=

7-^^—

^^^^.

=

95.5

(one metric

horse

(0.03

X

TT

X

1000)

power

=

75kg-m. per sec).



Ciur.

III.)

FRICTION

IS

STEP

BBARIN08.

28

Prob.

6.

—

Find

exp

rc

Mio

na eomqxmding

to

(5)

and

(9)

for the

work of frietkm

per

revohitkm for a

new

flat

step

bearing,

and aim

for

a

worn

atep

bearing,

when

the

shaft

is

boUow,

the

inside

and

the

outside

radii

being

ri

and

rt.

^«.

ir.^.|,(-;^;;:)

(y)

ir-Mi*-(r,

+

r,)

iV)

n

-

(solid shaft)

these expressioos

become

identical

with

(5)

and

(9).

Prob.

7.

—

Prove

(a)

by

reasoning,

(6)

nwthematically,

that the

work

of

friction

is

greater

with

a hollow

shaft than

with

a solid shaft,

provided

Uiat

the total weighty and the average

pressure per unit

area, are the

same

in

both

cases.

Solution.

—

(a) The

siq>porting area of

the

step'

bearing

being

the

same

in

both cases, the outside radius of

the hollow shaft

must

be

larger

than

that

of

the solid shaft.

Therefore,

the

average

path

described

by

elemratary

forces

of

friction during

one

revolution

is

greater

in

Uie

hollow shaft; consequently, with

a

similar

distribution

of

pr

eas

u

res, the woric

of

friction

is

greato*.

(6)

To

prove

that

the

value of

W

from

eacpression

(5')

is

greater than that

from

£q.

(5),

when

wt*

w

(r,*

—

r,*): We

have

ri

-

r,*

ft

+

r,

But

ft

>

r;

hence

rs-¥ri

A similar

proof

applies

to

Eq.

(9)

and

(9').

Prob.

8.

—

Prove

that the

work

of

friction in

a

perfectly-worn

step

bearing

is

expressed

by

IF

-

^rPr

•

2

r*r*.

Hint

—

Substitute

Eq.

(6)

into

(1).

Pkob.

9.

—

The

eaqxeesion

for

the

work of friction,

deduced

in

the

preceding

proUem,

eontains

the

unknown pressure,

p„

at

the

periphery of the

shaft.

Show

how

to

determine

this

{Measure when

the

variaUe ooefficient

of friction is

expressed

by |t

-

in

•

—

,

where

P

m,

m,

and

n

are giveo

coostanta; s

is the

linear

vdocity

at the

point

to

which

tt

refen.

Sohdion.

—

Vdodtgr 9 m

proportional to the

distance

of

the

point

from

the oenter, so

that Eq.

(6)

gives:

j»

•

—

•-

Constant,

or



24 ENGINEERING

MATHEMATICS.

[Chap.

III.

j-n+i.pi-w^

j.n+1

.p^i-m

Physical

conditions

are

such that

pat

the

center

is

always larger

than

the

average

pressure, and

theo-

retically

approaches

to

infinity.

Therefore,

the

foregoing

equation

has a

practical

meaning

only

when

m

<

1. Substituting th6

value

of

p

into Eq.

(3)

and

integrating,

results

in

P

=

pr

•

2

Trr*

•

_

_

.

•

From this

equation

the

unknown

pressure

pr

can

be

determined.

In

order

that

the

right-hand

side

of

this

equation be

positive

the

condition

must

be

fulfilled:

2m

+

n

<

1. When

m

=

n

=

the

expression

for P

becomes

identical

with Eq.

(8),

because

in this

case

M

=

Ml

=

Constant.

Note.

—

It

may

be objected

that

the

expression

for

fx

used

in

this

problem becomes

zero at

the center

of the

shaft

(v

=

0);

also

that

it

approaches to

zero

with

increasing

pressure,

and

approaches

infinity

with decreasing

pressure. A

better formula

for

/u

would

be

a

modification

of the

form:

/x

=

ah

[v

-|-

a]

-f-

1

-|-

b-.

—

f—

—

.

It

L (c

+

pr-i

gives

finite

values

for

m

within the range

of

pressures from

zero to

infinity; it

gradually

decreases

to

a

finite

value,

different

from

zero,

when

the velocity

v

approaches to

zero.

However, the

integration

with this

form of

expression

for

n

would

be much

more

involved;

moreover,

it

would be

hardly

warranted

in view of

the

meager

information

available

with respect

to

the actual

variations of

the

coefficient

of

friction

in

step

bearings.

Prob.

10.

—

Assuming

the unit

pressure to vary

inversely

as

the

n-th

power

of

the

distance

from

the

center

of

the

shaft,

prove

that

(2

—

n)

Pr

=

Pav

•

^^

•

Hint.

—

Substitute

the

value

of

p

from

the

ex-

p

pression

px

=

PrV^

into Eq.

(3).

The

pressure

Pav

=

-^-

When

w

=

1,

Pr

=

2

Pav,

&

rcsult

already

derived

in Prob.

4.

Prob.

II.

—

Deduce

expressions

equivalent

to

Eq.

(2)

and

(3)

when

the pivot,

instead

of

being flat,

is shaped

according to

a

sur-

face

of

revolution

(Fig.

8).

Solution.

—

Let

p

be

the

normal

unit pressure at

a

point

distant

X

from

the

axis;

consider

the

work

of

friction

upon

a

zone of

the

surface

having an

infinitesimal

width

ds.

Repeating

the reasoning

given

in

Prob.

1,

we

find

TF=4

7rV

Cpx^ds,

(10)

t/ro



Cbap.

Ill

1

FRICTION

IN

STEP BHA RINGS.

25

the

only

differenoe,

as

compared

with

Eq.

(2),

being

that

the

width

da

is

not

the same as the

increment

dx of

the

radius.

The

condition

<tf

equilibrium of

the

shaft

requires that the

sum

of

the

TOtioal

projeetiona of

pressures

p

be

equal to the

weight

P

of the

reviving

part, or

P-

I

pcos4'2rxd«

(11)

In order

to

be

able to

integrate

these

expressions, the

law

of

distribu-

tion of

{MresBures

must

be

known

and

the shape

of

the surface

of

revolution

given.

Fig.

8.

—

A

pivot,

shaped

according

to

a

surface

of revolution.

Prob.

la.

—

Simplify Eq.

(10)

and

(11)

for

the case

of a

new

bearing,

and show

that

the constant

normal

pressure upon the

sur-

face

of revolution

is

equal to the average

vertical

pressure

upon

the

projected

area

of the

pivot.

W'iw^^p,

I

x'di,

X

(12)



26


[Chap.

III.

where

the

constant

normal

pressure

po

is

determined

from the

relation

P

=

2irpo

j

X

cos

<i>ds

(13)

But ds

cos

<i>

=

dx, so

that

Eq.

(13)

integrated

gives

P=po'ir{r^-ro')

(14)

The

expression

tt

(r^

—

ro^)

represents

the

projected

area

FGHK

of

the pivot

(Fig.

8).

Let

the average vertical

pressure on this area

be

Pver-

We

have then P

=

Pver

'

ir

(r ^

—

r^),

and,

comparing

with

Eq.

(14),

see

that

po

=

Vver-

Prob. 13.

—

Apply

Eq.

(10)

and

(11)

to the

case of

a

perfectly

worn

bearing,

with

a

constant coefficient of friction.

Solidion.

—

Vertical

wear

w

(Fig.

8)

must

be

constant

at all

points:

but

w

=

,

where

n

is the wear normal

to the surface.

cos

<^

This latter wear

is

proportional to the work of friction

per unit area,

so

that

we

find,

as

in

Prob.

3,

that

n

is

proportional

to

the

product

vx.

Hence,

the

expression

—

—

must

be constant,

^

'

*^

cos<A

or

_P5_

=

.PrL

=B=

Constant,

(15)

cos

<t>

cos

o

where

B

is

the

numerical

value

of

the constant, and

«

is

the

angle

which

the

tangent

to

the

profile

at the periphery

of

the shaft makes

with

the

horizontal.

Substituting

the

value

of

px

from

Eq.

(15)

into

Eq.

(10)

and

(11)

gives

Tf

=2,rM5-7r(r2-ro^)

(16)

P=2irB

f

cos

4>ds, (17a)

or

P=2irB I

cos

<t>dx

(176)

In

applications,

either

Eq.

(17a)

or

(176)

is

more

convenient,

accord-

ing

to

the shape

of the

profile

of the

pivot.

Prob.

14.

—

Show

that, when

the

surface

of revolution

is an ordi-

nary

truncated

cone

(0

=

a

=

Constant),

the work

of

friction

is

expressed

by

^=(am.

4

(-:-)

08)

\cosa/

3

(r^-^ro)



Chap.

III.l

FRlCTlVS l\

STEP

BE

A

Rl

SOS.

iriieo the

bearing

b

new,

and by

I

Vcoso/

27

MM)

the

bearing

is

worn.

h. 15.

—

Show

that

Eq.

(5)

and

(9)

can

be

deduced

as

special

of

Eq.

(IS) and

(19);

also that

friction

Io«

is greater in a

oonieal

st^

bearing

than

in a flat

one,

with

the same

diameter

of

shaft

Prob.

16.

—

Show

that

the

woric

of

friction

in

a

new

conical

st^ bearing

\a

greater

than

in

the

same

bearing

after

it

has

been

perfectly

worn.

//

ml

.

J

(

r»

-

r.*)

-

3

(r»

-

r,«) (r+

r.)

-

(r

-

r,)»

>

0.

Fiff.

—

A

spherical

step bearing.

Prob.

17.

—

Prove

tliat

in a new

q>herical

step

bearing

(Fig.

9)

the wurk of frictiuti Ls

expressed

by

W

Irr-uP

a

—

Sin a

cos

a

mfi

(20)

//tn<.^Sub6titutez-A8in#.andd««

A(i#intoEq8.(12)

and

(13).

Prob. 18.

—

Show

that

in

a wmn

q>barical

step

bearing

the

work

of

friction

is

W

ring

+

ainaoo8<

2wr,

....

(21)



28

ENGINEERING

MATHEMATICS.

(Chap. III.

and that the unit pressure at

a

point defined by the central angle

^

is

expressed

by

P cot

4>

sin'

a

,nct\

P

=

r

•

^

(22)

wr a

+

sin

a cos a

Hint.

—

Use Eq.

(15)

and the

same substitution

as in the

preceding

problem,

into

Eq.

(10)

and

(11).

Prob.

iQ.

—

Eq.

(20)

and

(21)

must

become

identical

with

Eq.

(5)

and

(9)

when the

radius

R

of the sphere

becomes

infinite,

in other

words,

when a

=

0.

Show

that this is

so by

evaluating

the

indeterminate

form

-

,

which the equations

assume

when a

=

0.

Hint.

—

According

to

a

well-known rule of

the

calculus,

take

the

ratio

of derivatives so that

[

g

—

sin g

cos a

[

_

_

r

l

—

cos'

g

^1-

SJn'

g

|

sin'g

Ja=o

L

3

sin'

g

cos

g

Ja=0

^

r 2

sin' a H

^

r 2 n

^

2

Ls

sin'

g

cos

gJo

=0

l3

cos

gJa =o

3

A

similar

procedure

is applied to Eq.

(21).

Prob. 20.

—

Assuming

the

work

of

friction in

a

worn

flat

bearing

to be 100 per cent,

plot

a

curve

of

values

of

the work of friction

for

worn

spherical

bearings for the

same

shaft, the

radii

R

of

the pivots

varying

from

infinity

(flat bearing)

down

to

the

radius

r

of

the

shaft

(semi-spherical pivot).

Ans.

g

=

a

=

45°

g

=

90°

R=

oo

R

=

rV2

R

=

r

W

=

100

per cent

IT

=

110

per

cent

W

=

127 per

cent

Prob. 21.

—

In

the step

bearings

considered above, the

pressures,

which

are

evenly

distributed

when

a

bearing

is

new,

gradually

become larger in the

parts

nearest

to

the center

of the

shaft.

As

a

result,

the bearing

changes

its

shape,

and

wears

out

more quickly

than

it would,

if

the

pressures

could

remain uniformly

distributed,

in

spite

of

the

wear.

The problem

is

to

find such a

shape of the

profile of the

step

bearing

that the

normal

pressure is

constant

in

the new,

as

well

as

in the

worn bearing.

Solution.

—

Let

p

in

Eq.

(15)

be

constant;

then

the

equation

becomes

—

-^

=

Constant

=

T

(23)

cos

(t>

cos

g



Cbap. III.)

FRICTIOS

IN

STEP

BEAR SOS.

89

Thk

is the equation

of the

i^ofile

of

the

bearing.

The

eoutant 7

repranoti

the

leogth

of

the

tangent

(Fig.

10),

because

-^

b

the

00S4

length of that part

of the tangent to the

curve,

mchided

batwwu

the

point of

eontact

and

the ans of ordinates.

Consequently,

the

reqdred

curve

is

cle6ned

l^

the condition

that

the

lei^th

T

of

its

Fig.

10.

—

Schide's

pivot,

the

profile

lidng

a

trartrix.

tangent

is

oonstant.

litis [wqjerty

of the

curve gives

a

mm}^

method

of

construction.

Let

Ot

be

a

naall

weight

with

a

pencil

point fsstened

to

it

Let

OX and OY be

axes

of

codrdinatee

drawn

on

a

horiiontal

board, and let

the

pencil point

lie

originally

on the

ass

of

slisriwf.

Attach to

the

wei^t

a

{neoe

of

non-elastic

string

ahx

^

T

such

that

the end

6i

just

readies to the origin.



30

ENGINEERING

MATHEMATICS.

[Cdap. III.

Draw

the end

of

the string

along

the

axis

OY;

the weight

will

follow, and

the pencil will trace

the desired curve,

the tangent

being

at all

points

equal

to

the length

T of

the string. For

the

reason

of

this construction

the curve is

known as the tractrix.

The

complete

curve

consists of four

identical

branches,

which

begin on the

axis of

abscissae

at the distances

± T

from

the origin

and extend

up

and

down

approaching the axis

of

ordinates

asymp-

totically. Any part

of

such a curve can be used as

a

profile

for

a step

bearing,

because

all

parts

satisfy

the

condition

(23)

of

a

uniform

distribution

of

pressure

in

spite

of

the

wear.

Prob. 22.

—

A pivot, shaped

according to

the

solution

of

the

preceding

problem, is

known in

engineering

practice as Schiele's

anti-friction

pivot.

Show

that

the

work of friction in such

a

pivot,

new

or worn,

is

equal

to

W

=

nP'2irT,

(24)

and

determine which part

of

the

tractrix

should

be selected

for

the

profile

in

order

to

reduce the friction

to a

minimum.

Solution.

—

Eq.

(24)

is

derived

by

using

Eq.

(23)

and

the

con-

dition

p

=

Constant in

the

general Eq.

(10)

'and

(11);

these

latter

apply

both

to

new

and

worn bearings.

As shown by Eq.

(24),

the

work of friction

is

a

minimum when

T is

selected

as

small as

possible.

Eq.

(23)

and

Fig.

10 show

that

it is

best

to

select the part

of

the

tractrix

beginning

with point

Oi,

at

which

the

curve

is

tangent to

the axis of

abscissa?. T

is

then

equal to the

radius

r of the

shaft,

and

this

is

evidently the

smallest

value

which

can be

assigned

to

T.

The

inner

radius

ro

of

the

pivot

is

determined by the

condition

that

the pressure

p

must

not exceed

a

certain

safe

limit

indicated

by

experience, for

a

particular service for

which

the

shaft

is

designed.

Note.

—

A

comparison of

Eq.

(24)

with Eq.

(9)

shows

that

friction

in

Schiele's pivot

is

considerably greater than

in an ordinary

flat pivot; therefore

the

name

anti-friction

is

somewhat

a

mis-

nomer.

It

would

be more appropriate

to

call

Schiele's pivot

con-

stant-pressure

or

minimum-wear

pivot.

Prob.

23.

—

Transform

the Eq.

(23)

of

the

tractrix into one

in

rectangular coordinates

(Fig.

10).

T

T

Solidion.

—

X

=

T cos

<(>

Vx

+

tan..

^^_^^^J

from

which

-^

=

—

.

The

minus

sign

is

selected,

because

ax

X



Chap.

III.)

FRICTIOS

IS

STEP

BBARJNOS.

81

the

d«ivative

is

negative for

the

lower

right-hand

branch

of the

eiirve:ydeerea8eswheDxincreaees.

The btegral

of

this equation ia

y,TiM(T±:^ :f^-Vfrr?,

(26)

The

constant

c4

integration

*

0,

because

y

-

when

x

•

T.

For

purposes

of

calculation

this

equation can

be expressed

in

a

sim-

pler tonn

through

hyperbolic

functions,

since

tables

of such

funo

tioQs

are

readily

accessible.*

Namely,

(27)

..r|c«h-.(D-v/i-(f)|

...m

The

coordinates

of the

tractrix

can

also be ex

pr

e

sse

d

conveniently

through

the angle

^.

Substituting

the value

of

-^

«

cos

4

from

Eq.

(23)

into

Eq.

(25),

we get

^-^[Ln

tan(|

+

^)-8in4]

X

-

r

cos

.

The

curve

can

be

constructed

from

these

equations

by

assuming

different

values

of

4.

Prob.

24.

—

Show

that Eq.

(25)

and

(26)

can

be represented in

the

form^-/(^j;

this curve

may be

called

the

generalized

tractrix,

good for

any

value

of

T.

Plot

this

curve

for

values

ol

^

from

1.00

to

0.05;t

explain how

to modify

its

ordinatcs

and

ab-

sdsse

in

order to

cowftruct tlic

profile

of

a

Schiele's pivot, for

a

given

T,

and

to

a

cotain

required scale.

Atu.

|(

-

1.00

0.50

0.05

^

-

0.451 2.69

T

t

*

Hyperbolic

functions

are

being

used to

a

oonaiderable

extent

in

physios and

lately

in

engineering.

The

fundamental

properties of

these

funettoos

will be found

in

Ssaver's

Ma^tmaiieal

Handbook

(McOraw-

Hill),

and in

J.

McMahon's

Hifperbolie

FunetioHM

(John

Wiley

A Sons).

t

It

is conveoient

to

sdect

different

scales

for

abacissg

and for

ordi-

nat«i.

However,

hi

this ease the length of

the

taagent

is

no

longer



CHAPTER

IV.

CARRYING

CAPACITY

OF

BELTS.

A

CERTAIN

amount

of

mechanical

power

is

transmitted

to

the pulley

A

(Fig.

11)

by

means

of

the

belt

BC;Bis

the

driving

side

of

the

belt, and

C

is

the slack

side.

Fig.

11.

—

Belt

and

pulley.

The tension

Ti

on

the

driving

side

is

greater than

the

tension

T2 on

the

slack

side, the

difference between

the

two

being

the useful

effort

P

transmitted

to

the

pulley.

This

difference in

tension is

possible,

because

of

friction

between

the

belt and

the

pulley.

The

problem is to

find

the

relation

between

Ti

and

T2,

knowing the

32



Chap.

I\

\RRYING CAPACITY

OF

BELTS.

88

diameter

of the

pull(>\

.

the

angle

a of contact,

and

the

coefficient of friction

n

between

the

belt

and

the

pulley.

Consider an

infinitesimal

element

MN of

the belt

(Fig.

12)

corresponding

to

the central

angle

d^.

The

forces

acting upon this element

arr:

tension

T

opposing

T-^dT

Rg.

12.

—

Foroea acting

upon

an

infinit««imHl dornpnt

of a belt

the

motion,

a

slightly

larg<»r tension

T

-^^

dT

m

the

direction of the

motion,

and the

tangential

force of

friction

dF between the pulley and

the

belt.

Since

the element

of the belt

is

in

equilibrium, the force

of

friction

balances the

difference

dT of the

tensions T

and

T

-f-

dT.

This

fact

expressed

analytically

gives

the

differential

equation

for

the

solution

of the

problem.

Friction

between solid bodies

is usually

assumed

to

be

proportional

to the normal pressure.

In our

ra«<e,

tension

T

gives in the

direction

C'C a

normal

com-

ponent

Tsin

~

which

presses

the element

of the t)elt

to the

pulley;

tension

T

-S-

dT

gives

a

component



34


[Chap. IV.

(T

+

dT)

sin

-~

•

Hence,

the force

of

friction

upon the

infinitesimal

element is

dF=dT=

t,\T^in^^{T+dT)^in^.

.

(1)

Since

we consider an

infinitesimal element of

the

belt,

the

increment

dT

can

be neglected

in

the

last

term,

as

compared

to

the

tension

T

itself.

Moreover,

the sine

of

the

angle

~

can

be

replaced by the

angle itself

(length

of

the arc

of

radius unity).

Eq.

(1)

is

then

simplified

to

dT=nTd<t>,

.......

(2)

or

—

=

M

d<i).

Integrating

this expression over

the

whole

angle

of

contact, a,

gives

Ln|i

=

M«, (3)

from

which

f-;

=

^,

(4)

where

e

is

the base of

the

natural

logarithms.

The

useful

effort

exerted by

the belt

upon

the

pulley

is

P=T,-T,

(5)

Eq.

(4)

and

(5)

are

the

necessary

relations

connecting

the

tensions

Ti

and

T^

with

the other

quantities

in-

volved

in belt drive.

Prob.

I.

—

In

formula

(3),

angle a is in

radians

(why?);

trans-

form

the

formula

into

^^^1;=^'

......

(6)

where a

is in

degrees, and

n

=

0.27.

Ans.

Const.

=

^1^^

=

488.



Chaf.

IV.l

CARRYING

CAPACITY

OF

BELTS.

S6

W

Prob.

2.

—

Exffnm

the

tenaiona in

the

driviof

mde

and

b

the

•bek

aide of

the

belt,

is

functions

of

the

useful

effort

P.

An$.

Eq.

(4)

and

(5)

give

T,~P'-^

(7)

T,^p,--L-

(8)

Prob.

3.

—

A

belt haa to

transmit

a useful

effort

of 95

kg.;

the

MDf^

of contact

is

160

degrees,

the co^cient

of

friction

is 0.30.

What ia

the

trasion on the

driving

side, and

how

wide

does

the

belt

have

to

be, if

the permiaaible safe

tenaioo

ia 7

kg.

per

cm.

of ita

width?

Atu.

167

kg.;

24

cm.

Hint.

—

Use

formula

(7).

Prob.

4.

—

Check the

solution

erf

the preceding

problem by

using

directly

Eq.

(5)

and

(6).

This

method

has

aome

advantage

because

it

is

Doi

oeceasary

to

evaluate

the

eqMmential

expreesion

e**.

Note

that

the

conatant

in

formula

(6)

must

be

modified

to

suit

the

pven

value of the

coefficient

of friction.

Prob.

5.

—

When

one

has to calculate

regularly

a large nuiiUHr

of

bdta,

using

Eq.

(7),

it

is

convenient to

prepare a table

or plot a

curve,

giving

the values

of

the

exponential

expression

by which

P

is

to

be

multiplied

in order

to

obtain

Tt.

Plot such

a

curve

for

M

-

0.28 and

for angles

of

contact

from

90 to

270

degrees.

Ana.

a

- 90°

180°

270

^

-

2.81

1.71

1.36

Prob.

6.

—

Elxpress

Eq.

(7)

and

(8)

through

hyperbolic

funo-

tkms.

Solution.

—

s>*

g***

_

Coeh

>

^

-f

Sinh

j

tta

*--

I

e***

-« **

2

8inhJ>ia

80

that

T,'~{Q:oihkm+l)

(9)

r.-^(CothJ^-l)

(10)

Prob.

7.—

Cheek

the

aolutitm

of

Prob.

3

by

means

of for-

mula

(9).

Prob.

ft.

—

When

the

periphefal

apeed of the belt

ia

oonaideraUe,

oeQtrifugal

foree

acting

upon

it

tendi to

aqwrate

it

from

the

pulley.

Analogoualy,



36

ENGINEERING

MATHEMATICS.

[Chap. FV.

Consequently, the

normal pressure, and

the

friction

between

the

belt

and the

pulley,

are

reduced,

as

well

as the maximum

eflFort

which the

belt can transmit without slipping.

Deduce

a

relation

between

Tx

and T^,

analogous

to

Eq.

(3),

but in which

centrifugal

forces

are

taken

into

account.

Solution.

—

According to the general

laws

of circular motion,

^

.,

,

-

mass

of

the

body

X

(velocity)*

t ^ i. ^i.

centrifugal

force

=

,.

, /

-•

Let

w be the

radius

of rotation

weight of one

meter

of belt, in kilograms;

then the

centrifugal

force

acting

upon

the

element

of

the

belt,

corresponding

to

angle

I

w

\ f* w

diff

(Fig.

12),

IS

[~

'rd<t>]

'-

=

-

v^dtk,

where

g

=

9.81

is the

\^

I

r

g

acceleration

due to

gravity (in

meters

per sec.*), r

is

the average

radius

of

curvature

of the

belt

on

the

pulley, in

meters,

and v the

linear velocity

of

the

belt,

in

meters

per

second.

This

centrifugal

force

reduces

the

normal

pressure

upon the pulley,

so that

Eq.

(2)

becomes

This equation,

after

being integrated

similarly

to

Eq.

(2),

gives

Ln

(

J,

_xm^\

rp

_

wv

=

/ua.

(11)

When

the

speed

is

low

so that the

correction term

—

can

be

g

neglected,

Eq.

(11)

becomes

identical

with

(3).

Prob.

Q.

—

Deduce formulae analogous to

(7)

and

(8),

taking

the

centrifugal

force

into consideration.

Ans.

Writing

Eq.

(11)

in the exponential

form

and

solving

with

Eq.

(5),

we

get

„na

,,,,.2

T,

=

P.-^

+

—

,

(12)

e^-1

g

and

1 o„„2

(13)

Prob.

10.

—

Solve

Prob.

3,

taking

the

centrifugal force

into con-

sideration.

Assume

v

=

16 m. per

sec,

and the

unit

weight

of

the

belt

=

7.3

kg.

per

sq.

m.



Chap.

IV.J

CARRYING

CAPACITY

OF

BELTS.

91

Solution.

—

Lei

the unknown

width

of

the

belt

be

x em.;

then

I

taorioD

7*1

•

7

X,

and

the

weight

per

meter

of

ienirth.

w

-*

-i^

•

100

Substituting

these

values into

Eq.

(12)

and

using

the

answer

of

Prob.

3,

we obtain:

7

x

-

167

+

1.9

x,

whence

x

-

32.7

cm.

Ten-

sion r,

-

7

X

-

229

kg.

Prob. II.

—

With

what

original

tension

must

a

belt

be

put

on

its]

pulleys

in

order

to

transmit

a

uadul

effort

P

?

Disregard

the

action

of the

centrifugal

force.

Solution.

—

The

original

teii.sion

To

on

either

side of

the

belt

is

an

arithmetical mean

of the

working

tensions

Tt

and

Tt,

so

that

from

Eq.

(7)

and

(8)

•

2

(<f--l)

^ '

or,

in

hyperbolic functions, according

to Eq.

(9)

and

< 10),

T,-|.Coth§po

(15)

Note.

—

The sum

of

the

toisions

Ti

and

Tt

remains

constant

with

variations

in

load,

and

is

equal

to

the total

original

tension 2 T,.

This is because

t«Dai(»s

are proportional to elongations,

and

the

extra

elongation on

the

driving

part

must

be compensated

for,

by

a

reduced elongation on

thn slack

side.

Prob.

13.

—

What

original t* usion

should

be given

to

the

belt

considered

in Prob.

A,

in order

that the

belt

could stand

2o

per

cent

ovorkNul,

without slipping

on

the

pulley ?

Am.

About

150

kg.

Prob.

13.

—

C] orrect

formula

(14)

for

the effect

of

centrifugal

force.

An,.

T..i(r.

+

r.)-?-^^;+H:.

. . .

(,6)

This

result

shows

that,

on

account of

centrifugal

force,

it

is

ncces-

sary

to

give

the

belt

a

larger original tension.

When

running

at

a

speed

V

the

part

—

of

Um

tension is

balanced by the

centrifugal

foree,

and only the rest

is effective in

producinir

friction

upon

the

pulley,

with

a

consequent turning effort.

Prob.

14.

—

What

should

be

the

original

tension

of

the

U'lt in

Prob.

10,

in

order

that

it would not

slip at 2.5

]>er

cent overload

?

An.. 150

+

?A>' ff-><li'-212kg.

v.oi



38


[Chap.

IV.

Prob.

15.

—

The

belt

considered in

Prob.

10

is

put on its pulleys

with a

tension

of

200

kg.

per side, at rest. Plot a curve

showing

the

maximum

useful

effort

P

which

can be

transmitted

with

this

belt at

different speeds. Determine the critical

speed at which

the

centrifugal

force balances the whole

of

the original

tension,

so that

the belt does not press

upon

the pulley, and therefore

exerts no

useful

tangential

effort.

An,. Eq.^16)

becomes:

f-'^ -

Off' ''

The

following points

lie

on this

curve:

v=

10 20

28.7

P

=

158 139

81.3

The

last

point corresponds

to the

critical

speed.

Prob. 16.

—

Plot on the

same

curve sheet

with

that

of the pre-

ceding

problem,

a

curve

of the

useful

power,,

in

kilogram-meters

per second,

transmitted

by

the belt.

Ans.

Power

=

Pv,

so that

when

y

=

10

20

28.7

Power

=0

1390

1626

Prob.

17.

—

The result of the preceding problem

shows

that the

power

transmitted by the belt reaches its maximum

at

a certain

speed,

and

decreases with

a

further

increase

in

speed,

becoming

zero

at

the

critical

speed. Show that the speed corresponding

to

maximum

power

is

equal

to

—

=

of

the

critical

speed,

and

that

at

v3

this

speed

one-third

of the

original

tension is

taken up

by

the

cen-

trifugal force.

Hint.

viTo

}

reaches its maximum

when

t,

=

i/l^;

Critical speed

=

l/—

Prob.

18.

—

Check

the

point

of

maximum

power, obtained

in

Prob.

16,

with the

formula deduced in

Prob.

17.

Ans.

1747

kg.

at

16.55 m. per

sec.

Prob.

19.

—

Supplement the curves obtained in

Prob,

15 and

16 with curves of tensions

Ti

and T^.

Ans.

V

=

10 20

28.7

ri=279

269.5

240.6

200

7^2=121

130.5 159.4

200



XSmat.

Vr.\

CARRYING

CAPACITY

OF

BBLTS.

89

Prob. ao.

—

The

results

of

the preceding

proUem

flhow

that

the

tensioD

Ti

deow

a

se

s,

while

T,

inereMes with

speed, when

the

hett

tnnsouts

its

nugdmum

tractive

effort;

at

the

eritieiJ

qpeed

7*1

.

Tt

-

r*.

Prove these

results in

the (eoeral form, and

shofw

that

the

curves

of

7*1

and

Tt

are parabohe,

one

turned

downward,

the

other

upward.

Hint.

—

EUniinate

P

from

Eq.

(12)

and

(13)

by

of

formula

(16),

and use

the

critical

qpeed

from

Prob.

17.



CHAPTER V.

TORSION

OF

SHAFTS.

Some

mechanical

power is

transmitted

from

pulley

A

(Fig.

13)

through

the shaft

B

to

pulley

C.

The

pulley

C

resists

the rotation,'^so

that

the

shaft

is subjected to

two

equal

and

opposite turning

efforts,

those of

A

and

of C; hence the shaft

is

twisted,

or,

in engineering

^

earing

^B

Shaft/

Pulley*

Belt

^

Fig.

13.

—

A shaft

and

two

pulleys.

language,

is

said to be

subjected to

torsion.

It

is

required to

find

a

relation

between

the

turning

moment

acting on the

pulleys

and the

safe

size

of

the shaft.

For

the

purposes

of

mathematical analysis,

the twist-

ing

of

the

shaft

is considered

with

reference

to

its

con-

secutive

cross

sections.

Each

cross

section is turned

by

an

infinitesimal

angle

relatively to the next cross

sec-

tion,

bringing

into play the

elastic reactions

of

the

material.

According

to the

fundamental law

of

equi-

librium,

the

reaction is equal to

the action, so

that

the

resultant

moment

of

these

elastic

forces

in

any

cross

40



Chap.

V.

TORSION

OF

SHAFTS.

41

Fig.

14.

—

CroB8-«ection

of

a

shaft

subjected to

torsion.

section

of the

shaft

i»

equal and

opposite

to the turning

moment

applied

to

the driving

pulley.

This law

makes

possible

to

establish

a

relation

between

the

applied

ment

and

the

elastic

stresses

in

the material.

Let

Fig.

14

represent

a

cross .section of

the

shaft,

and

let

9

be the

resisting

force

in

the

material,

in

kilograms

per one square

centimeter

of

cross section,

at

a

distance

of

X centimeters

from

the center

of

the

shaft.

This

unit

force

is

called

the shearing

stress in

the

material,

since

it

is

caused

by

the

tendency of one

cross

section

to slide

with

respect

to

the

next

cross

section.

The

shearing

force

upon

an

infini-

tesimal

annulus

of

the

width

dx

\»

q

'

2

rx

•

dx,

and

the

moment of

this

force,

with

respect

to the

axis of

the

.shaft, la

q

•

2

rx

•

dx

•

x.

The

above

stated

condition

of

equilibrium

becomes, there-

fore,

M''2xrqx'dx

(1)

where M

is

the

applied

turning

moment.

In this

equation

q

is

& variable

quantity,

so

that,

in

order

to

perform

the

integration,

q

must

be

given

as

a function

of

the

distance

x

from

the

center.

According

to

the

fundamental

assumption

of the

theory

of

elasti'

^sses

are

proportional

to the

-ponding

strain-

or

di.splacements

(as

long as

a

Lin

limit,

called

tho ohistio limit

of the

material,

is

not

exceeded).

In

our

case, assuming

that

each

rro.'^s

section

turns as

a

whole

with

respect to

the

next

cross

section,

displacements

an*

th««

lMrp<*st

at

Ww

periphery,



42

ENGINEERING

MATHEMATICS.

[Chap.

V.

and

decrease

to

zero

at the

center.

The angle

of

turn-

ing

being

the

same

for

all

the

points

of

a

cross

section,

actual linear

displacements

are

proportional

to the

dis-

tances X from the

center.

We

have

thus, remembering

that

stresses are

proportional to

displacements,

^

=

^,

(2)

qr

r

where

qr

is

the

stress at the periphery

of

the

shaft.

This

stress

is larger

than the

stress

at any

other

point

of the cross

section, so that

if

qr

does

not exceed the

safe

limit,

determined

by

experience,

the whole

shaft

is

safe with

respect

to torsion.

Substituting the

value

of

q

from

(2)

into

(1),

we obtain

r

Jo

or

M

=

^ —^

(3)

This

is

the

required

relation

between

the

turning

moment

M

and the

maximum

stress

qr

in the shaft.

If

qr

is in

kilograms

per

square

centimeter, and

r

is in

centimeters,

the

applied

moment

M

must

be

expressed

in

kilogram-centimeters.

Prob.

I.

—

Draw

a

curve

giving

values

of safe

twisting

moments

for

shafts

up

to 30

cm.

in

diameter; assume

the safe

permissible

stress

to

be

600

kg. per

sq.

cm. What

is

the

mathematical

name of

the

curve?

Ans.

M

=

31,800

kg-m. for

D

=

30

cm.;

cubic parabola.

Prob.

2.

—

The

diameter

of

the

shaft

can

be expressed

from

formula

(3)

as

D=h'^M,

(4)

where

ki

is a

constant.

Calculate

ki

for

qr

=

800

kg.

per

sq. cm.,

so that

D be

expressed

in

centimeters when M is in

kilogram-

meters.

--

*-/^l^=««^-



Chap.

V.l

TORSION

OF SHAFTS.

48

Prob.

3.

—

In

practice,

the

power

P

txanamitted

through

the

shaft,

aod the

speed

N

of nation are

usually

given

instflad of the

turaiog

moment

M.

Elxpression

(4)

becomes,

then,

D'kt^'~,

(5)

where

kt

is

another

eoostant.

Express

Art

through

ki,

when

F

is iu

kilowatts,

and A'

in

rev.

per

min.

(1

kg-m. per

sec

«

9.81

watts.)

Ans.

P._..V.—

.1.02

—

hence

kt

-

.^

10

*,

(nearly).

Vim

Prob.

4.

—

How

is

formula

(3)

modified

when

the shaft

is

hoUow,

the

inside

radius

being r«?

Solution.

—

M

-'-^

r

I'dx-^'ir^-U*)

(6)

or

A'-^r-y-a-n*)

(7)

where

n

-

-

.

For

a

solid

shaft

n

-

0,

and

Eq.

(7)

becomes

identi-

r

cal

with

(3).

Prob.

5.

—

A

hollow

shaft

has

the

inside

diameter

equal

to

00^

half

of

the

outside

diameter.

What

is

the

saving in

material, as

ooo4)ared

with

a

solid shaft

having

the

same

outside

diameter,

and

what

is the

corresponding

loss in

strength?

An$.

Saving

in material,

<

-

n*

-

0.25

-

25

per

cent; loss in

safe

resistance

to

torsion,

/

-

n*

«

0.0625

-

6.25

per cent.

Hint.

—

The

volume

of

a

hoUow

shaft

per centimeter

length

is

expressed

hiy

r(r«

-r,»)-»r«(l

-n«)

(8)

Prob.

6.

—

ExtMid

the

solution

of

the

preceding

problem to

values

of

n

from

lero to

unity,

and

plot

the

results

in

the form

of

curves,

against

values of

n

as

iiliiini—

1.

Ant.

n

9

I

0.5

25

6.25

1.0

100

100



44

ENGINEERING

MATHEMATICS.

[Chap.

V.

Prob.

7.

—

Referring

to

the

preceding

problem,

at what

value

of

n

is

the

difference

between the numerical values

of

s

and

I a maxi-

mum?

•v/2

Ans. n

=

—

^

=

0.707.

Hint.

—

Equate

to zero the

first

derivative

of the expression

(n2

-

n').

Prob. 8.

—

Show that the

solution

of Prob.

7

corresponds

to the

value of n

at

which

the

tangents

to the two curves (Prob.

6)

are

parallel

to

each

other.

Prob.

9.

—

Two shafts

are

given,

of

equal

length

and

weight,

and

are made of

the

same material.

One

shaft

is solid, the

other

shaft

is

hollow

and

has

the

inside

diameter

equal

to

two-thirds

of

the

outside diameter.

How much stronger is the second

shaft?

Draw the cross sections

of

the

two

shafts

to

the same scale, for

a

comparison.

Ans.

1.94

times

stronger.

3

The

ratio

of

the

outside diameters

is

—

z

=

1.342.

V5

Prob.

10.

—

Find

the value

of n

at

which

a

hollow shaft

has

the

greatest

resistance

to twist, per

unit weight.

Solution.

—

From Eq.

(7)

and

(8)

we

see

that

the ratio

^Tr^-^<^

+

'>

<^)

must

be

a

maximum.

This

expression

increases

indefinitely

as

n

and r increase

;

the

maximum value of

n is unity,

and

the maxi-

mum

value

of r

is

infinity,

so

that

theoretically

the

shaft

of maximum

resistance

per

unit

weight

is

a

tube

of

an

indefinitely

large radius,

with

infinitesimal

ly

thin

walls. This

result

could

have

been

foreseen

from

the

theory of

torsion, given

above;

only

the material

at the

periphery

of the shaft

is stressed

to

its safe limit,

so

that, in order

to

get

the

full advantage

of the

material,

all

of it

must be

concen-

trated at the

periphery,

or

as

near to

it

as

possible.

Moreover,

the

periphery must

be as

far

as

possible

from

the

center, in

order

to

increase the

resisting

moment of

the

stresses

with

respect

to

the

axis

of

the

shaft

(see

Fig.

14).

Hence,

in practice, hollow, thin-

walled

shafts,

of

as

large a

diameter as

is

feasible,

approach the

condition

of

maximum

strength with

a

minimum weight.

Prob.

II.

—

Let the

requirement

of

the

preceding

problem

be

limited

by

the condition

that

the

thickness

of

the

wall

of

the

hollow



Chaf.

V.|

TOR^lo\

f>F

SHAFTS.

4'»

Atdi

must

be

not less than a

certain

practical

limit

of t centi-

metera.

Is

the

result

mocUfied

thereto?

Solution,

—

The

new

condition

is

ex

p

ress

e

d

by:

r

(1

-

n)

«

(.

Substituting

this

value

of

r into

Eq.

(9),

we find

that the

egprm

iOD

-^»

<•«>

IK

Subst

L

be a

maximum.

This expression

increases

with

n,

and

readies

Ha largest

numerical

\'alue,

equal to infinity,

at

n

-

1.

But

n

-

1,

with

a

finite

thickness

oS

the

wall,

means that

the radius

of

the

shaft must

be

infinite.

Consequently,

a

hoUow

shaft,

with

a

given

thickness

of

the

wall, is

the

strong^'

per

unit

weif;ht,

the

larger it«

radiuM.

This

result

could

have also

been

forcMccn directly

from the

theory of

torsion.

Prob.

I

a.

—

Suppose that the

student did

not

see directly

that

y

in

Eq.

(10)

reaches

its largest

numerical

value

at

n

-

1.

Pro-

ceeding,

then,

according

to

the

rule of

the calculus,

tliat

is

to say,

equating

the

first

derivative

of

y

to

zero,

he

would

find

two

solutions,

rtf.,

»

-

1

±

V2.

Plot

cur\'e

(10)

Ijetween

the limits n

-

-

I

and

n

•

+ 3,

and

explain

the meaning

of

the

thrpe

solutions

for

n,

determining,

if necessary,

the

sign

of the second derivative.

Prob. I

J.

—

A

hollow

shaft

has

to

withstand

a

given

torsional

moment and

be

as

light

as

possible.

.Find Wa external

and internal

radii.

Solution,

r'

(1

—

n*) must

be

a

miniinuni,

with the

Umiting

con-

dition that

t*

(l

—

n*)

is

constant.

This

latter condition

can

be

written in

the form:

r»

•

r

•

(1

-

n*)

(1

-f

n*)

-

Constant,

so

that

•

#. a^

Constant

.

.

:

+

n')

-

maximtmi.

This

is

identical

with

the

condition

(9),

su

that

the

solution

of

Prob.

10

applies

in

this

case.

Prob.

14.

—

The

cost of

making

a

shaft increases

with

its

diameter.

eq)eciaUy in

large

sixes;

therefore the

weight

is

not

the on

I

wfaieh detamines

the

cost

<^

the AtJi. Only the weight

ha

taken

into

eoosidention

in Prob. 10 to

13.

In order to

.v

.

itow

the

cost of

manufacture

can

be

taken

into

accoimt,

assume

tirnt

the

price for

rolling, per centimeter of

length, is

expressed by

the

formula

a

+

br

cents,

where

a

and

6

are empirical

constants. The

prioe

of

steel if

p

oenta

per

cubic

centimeter. Fmd

under

these

eonditkiQS

the

dkntndnim

of

a

hollow

sliaft

which

must

stand

a



46

ENGINEERING

MATHEMATICS.

[Chap.

V.

given

torsional

moment,

and

the

cost

of

which,

per

unit

length, is

a

minimum.

Solution.

—

Eq.

(7)

gives

r»

(1

_

n*)

=

—

=

Constant

(11)

Trqr

The

total cost

of

making

the

shaft,

per

centimeter

of length,

includ-

ing

the cost

of

the material, must

be

a

minimum,

or

P'lrr^

{I

—

n'')

+

a

+

br

=

minimum.

. .

(12)

The

usual

method

of

solving

equations

like

(11)

and

(12)

consists

in

eliminating

one

of

the

independent

variables,

n

or

r,

and

differenti-

ating

with

respect

to

the

other

variable.

It

is

simpler,

however, in

this

particular

case, to differentiate

both

equations first.

Namely,

the complete

differential

of

expression

(11)

is

equal

to zero,

because

the expression

itself

is

equal

to a

constant;

the

complete

differential

of

expression

(12)

is

equal to zero for

the

values

of n and r which

convert

it

into a

minimum. Thus we

have

3

r^dr

(1

-

w')

-

r^

•

4n'

dn

=

0;

2r

'dr (I

-

n^)

-

r^ '2ndn

+

b'

dr

=

0,

where

a

new

constant b'

is

introduced for

the sake of

abbreviation

b'

=

—

(13)

irp

Eliminating

dn and

dr

from the

foregoing

two

expressions

gives,

after

reduction,

2^

=

4+n^-4

(14)

The

required dimensions

of

the

shaft

are

found by

solving together

Eq.

(11)

and

(14).

Prob.

15.

—

Apply the solution of the preceding problem

to

the

following practical case:

A

propeller shaft

for

a

large

steamer

must

be

designed

for

a

maximum

twisting

moment

of

255,000

kg-m.,

the

limiting

stress

qr

not

to

exceed

250

kg. per

sq.

cm.

The

price

of

steel

is

40

dollars

per

metric

ton,

and the

rolling mills

charge

6.325

dollars

extra,

per

meter length

of

shaft,

for

each

centimeter

of

in-

crease

in

diameter,

above a

certain

size.

Solution.

—

The constant in

Eq.

(11)

has the

value

of

2X255,000X100

^.f.^^,

~-r

=

64,9C0cu.

cm.;

2w0ir



I

COAP.

V

TORSIOS

OF SHAFTS.

47

.

6.325

X

lOQ

,o«s »

1-

.1

Qc

w

iQQ

^2.65

oenUi

percm.

radiiw,

per

cm.

length

40x100x7.8

n/»io

___„^.

.0.0312oent.percu.cm..

where

7.8

is

the

specific

gravity

of

steel.

Therefore,

.,

12.«5

,^

*-d:03T2T-'29«q.cm.

The

values

of

n

and

r

are found

cither by

triala,

or

by

plotting

the

curves

represented by

Eq.

(11)

and

(14),

and

determining

the

point

of

their intersection.

Aru. r

-

41.5

cm.;

n

-

0.55.

Prob.

i6.

—

Check the

solution

of

the

preceding

problem

by

plotting

curve

(12)

of the cost

of the shaft. Take

values of

n

between

sero and

0.90,

and the

corresponding

values of

r

found

from

Eq.

(11).

For

the

same values

of

r and n,

draw

also

a

curve

of

the

weight

of

the

shaft

per centinwter

length.

Plot

both

curves

on the

same

sheet,

against

values

of

n

as

abedaBS.

Aru.

n,

%

r,

cm.

Weight, kg. Cost,

dollars

40.19 39.59

6.68 +

55

41.50

29.43

6.40

+

a

(minimum)

90

57.36 15.30

7.87

+

NoU.

—

The curves

show that,

in

spite

of

wide

differences in

weif^t,

the total

cost

remains

practically

constant

within

a

wide

range

of

suitable

values for n

and

r.

This

is

because

the

saving in

material

with

larger radii is nearly

compensated

for

by

a higher

cost

of

rolling.

Numerous

other

conditions,

which

usually

enter into

any

engineering

problem,

reduce

this

range

of

radii

to

much

narrower

limits.



CHAPTER

VI.

MOMENT

OF

INERTIA

OF

FLYWHEELS.

In

many machines, such

as steam

and

gas engines,

pumps,

punches,,

etc.,

the effort on the

shaft

varies

periodically

within

wide

limits.

At

the

same

time,

the

speed of rotation must

be

kept constant within narrow

limits. This is accomplished by mounting on

the

shaft

a heavy wheel

(Fig.

15)

—

so-called

flywheel

—

which,

on

Fig.

15.

—A

portion

of

a

flywheel.

account

-of its large

inertia, tends

to

smooth fluctua-

tions

of speed.

At moments

when the driving

power

exceeds

the

power actually

spent

in

useful

work

and

48



fM

C»ir.

VI.)

MOM

EST

OF INBRTIA OF FLYWHBBLS.

49

friction,

the surplus

of

energ>'

accelerates

the machine,

including

the

flywheel.

But

it

takes

a

considerable

amount

of

energy

to accelerate

the heavy

flj-wheel;

therefore,

the

same

surplus

of

energ>' changes

the

speed

of

the

machine

by

a

smaller

amount

than it would

out the

flyifv'heel.

Wlien,

at a next moment,

the

driving

power

falls

below

the

required

powTr,

the

machine

and

the

flywheel

slow

down,

giving

up a

part

of

their

stored

energj'.

Here,

again, a

comparatively

small

reduction

in

speed

Is

sufficient to liberate

a

<

on-

siderable

momentum,

stored in

a

heavy flywheel.

In

brief,

the

flj-wheel

acts as

a

storehouse

of

energy,

periodically

absorbing

and

releasing it

through small

variations in

speed.

It

must

be clearly understood,

however,

that

a

flj'wheel

Ls

not

in

itself

a

source

of

energy;

therefore,

the

average

driving power must

be

equal to

the

average

power

demanded

of

the

machine,

otherwise

the

machine

will either

gradually

speed

up

or slow

down

in

spite

of the

fl>'wheel.

The

dimensions

of

a

flywheel are

determim'(l

hy

tin'

condition

that,

with

a

given

range of

fluctuation

of

power,

the

variations

in

speed

should

not exceed a

given

limit.

It

is

therefore

necessarj'

to

know

how

to

determine

the

amount

of

energy-

stored

in a

flyijvheel

of certain

dimensions

and

weight,

revolving

at

a

given

speed.

Theorj' and

experiment

show

that the

energj*

pos-

sessed

by

a

material point

of

weight

p,

moving

at

a

velocity

«,

is

?^

.

where

g \»

the

acceleration due

to grav-

i

\

If

p

is

in

kilograms,

and

the speed

a

in

meters

per

second,

the

energy

is

in kilogram-meters;

the

accelera-

tion

of

gravity

g

-

9.81

inetiTS

per sec*.

The

foregoing

expiession

for

energy

is

true

for rectilinear

motion

and



50

ENGINEERING

MATHEMATICS. [Chap.

VI.

for

a

circular path.

Hence,

the energy

possessed

by

a

flywheel

is

^-hf'''

^'^

where

dp

is

the weight

of

an infinitesimal

element

of

the

flywheel,

moving

at a

velocity

s. The

integration

is

extended over

the whole

volume of

the

wheel.

Expression

(1)

can

be transformed

as

follows:

Let

the flywheel

revolve

at

a

speed

of

A^

revolutions per

minute; then

the velocity of

a point,

distant

x

meters

from

the

center,

is, in

meters per

second,

^=2«i

(2)

Further,

if

pi

is

the

weight

of

one

cubic

meter

of

the

material of the flywheel,

we

have

also

dp

=

pidv,

(3)

where dv is the

corresponding

element of

the

volume.

Substituting

(2)

and

(3)

into

(1)

gives

-=f.-(¥)

•-/-*

• • •

(^)

Apart from

the

constant multiplier

—>

this

expression

consists

of two factors,

one depending only

upon

the

speed

of

the flywheel,

the other

only

upon its

shape

and

material. This

latter

factor, which

properly

characterizes the

flywheel

itself, with

respect

to

its

capacity

for

storing

energy,

is

called its

moment

of

inertia.

Thus,

Moment of

inertia

=

pi

I x^dv.

...

(5)

Jo

Since

pi

is only

a

constant

by

which

the

results

of inte-

gration

are

multiplied,

we

shall

confine

ourselves



Chaf.

VI.)

MOM

EST

OF

ISBRTIA

OF

FLYWHBBLS.

61

chiefly

to

the

geometric

moment

of

inertia,

denoted

here

by

/.

That

is

/»

/

x«dr

(6)

expressed

in

meters

to

the fifth

power

(m.^),

when

the

volume

is

in

cubic

meters,

and

a; is in

linear

meters.

The

object

of

this

chapter

is

to

show

how

to calculate

/

in

some

important

practical

cases.

Prob. I.

—

A

hollow bra^ cylinder,

with

walls

5

nun.

thick,

*nd of

an average radius

0.5

m., revolves

around

its

geometric

axis,

at

a

qwed of

100

rev.

per

min.

How

much

enerfcy

is stored

in

the

e^dinder,

per

centimeter

oi

its

lenKth? S^3eci6c

gravity

of hnm

is

8.55.

Neglect the

small

amount

of

enerj^r

stored in

the light

spokes

which

fasten

the

cylinder

to

the

shaft.

Solidion.

—

The

thickness <A the

cylinder

is

so small,

as

com-

pared

with

its

radius, that

the

whole volume

can

be

assumed, with-

out

an ai^xeetable error,

to

be concentrated

at

the average

radius

of the

cylinder.

In

other words,

x

in Eq.

(6)

can

be

taken

as con-

stant

and equal

to

0.5

m. The

volume

of the cylinder

per centi-

meter

length

is \-er>'

nearly

equal

to 2 »

x

0.5

X

5

X

10 *

X

10 *

-

15.706

X

10~*

cu.

m.

Consequently,

the geometric moment

of

inertia

/

-

0.5'

X

15.708

X

10-*

-

3.927

x

10-»

m».

Then,

accord-

ing

to

Eq.

(4),

^

(

I9W)

^

^C^m^y^

(1000 X 8.55) X

3.927

X

10-

•

1&77 kg-m.

Prob.

3.

—

Deduce from the

solution

of

the

preceding problem,

a

general expression for the

moment

of

inertia

of

a

thin

cylinder.

An*.

/-2w6<.r,»,

(7)

wh«e

(

is

the thickness of the rim,

6

its

width,

and

r»

the

average

radius (Fig.

15).

Prob.

3.

—

Fmd a

more

general ex

p

ression than

the

one

given

above,

for

the

moment c^

inertia

of

a cylinda , when its

thickness

I

is

not

small

as compared with

the

averafe

radius.

Sofufum.—

The

«]iinder

can

be

eon

ri

dered

as eouisttng

of

infinitely thin oonoentrie shells; the volume

of

an

infinitesimal

shell

b

dp

•

2

rx

•

dr

•

6,

where

6

is the

wkith

of the

cyiioder,

paralJd



52


[Chap.

VI.

to

the

axis. Substituting

this expression

for dv into

Eq.

(6)

and

integrating,

gives

I

=

2irb rx^dx

=

'^-{r*-ro*),

....

(8)

where

r is the outside

radius,

and

ro

the inside

radius

of the cylinder.

Prob.

4.

—

Show that with the data

given in Prob.

1,

the

moment

of

inertia calculated from

formula

(7)

is

practically the

same as

cal-

culated

by

means of the

more

accurate

expression

(8).

Note.

—

In applying formula

(8),

use tables

of

logarithms (not

a

slide

rule)

and

be

very

careful

about

the

last

decimals.

Ans.

/

=

^

•

(0.5025

-

0.4975

).

Prob.

5.

—

The

answer

to the

preceding

problem shows

that

formula

(8)

is

inconvenient for numerical

computations when the

thickness

of the

cylinder is small

compared

with

its diameter.

Express

(8)

through

the outside

radius

r and

the thickness t of the

rim,

so

as to

avoid

computing

a

difference

of

two

quantities

r*

and

ro*

nearly equal

to

each other.

Solution.

—

ro

=

r-t;

(9)

hence

r*

-

ro*

=

r*

-

r*

+

ir't

-

6 rH^

+

4

rf

-

<*;

or,

after

reduction,

/

=

2.6^r'[l-1.5Q+(^)'-0.25(^y]..

.

(10)

The

terms

within

the brackets

converge

rapidly with

small

ratios of

^

;

in

most

practical cases the

first

two

terms

are

the

only

ones

to be

r

considered.

Prob.

6.

—

Check the

solution

of

Prob.

4

by means

of formula

(10),

and

show

the

advantage

of

the

latter in

numerical

computa-

tions.

How

many terms

in formula

(10)

are

needed in

this

particu-

lar

case?

Prob.

7.

—

What

is

the

largest

value

of the

ratio

-

with

which

the

r

last two

terms

in

formula

(10)

can

be

omitted, provided

the error

must

be

not

over

one per cent?

Solution.

—

From

the

condition

(?)'

=0.0.[l-,.5(^)

W?J

],

we

find

(-

1

=

0.0932.



Cbap.

VI.|

MOM

EST

OF

INERTIA

OF FLYWHEELS.

68

Prob.

8.

—

Extend

the

lolutioD

ci

the

pneedii^

problem to

values

of error

from

lero

to

10 per

eeot,

and

plot

the reeuHs

in the

form

of

a

cur\-o.

y/S?-,.,

,,,

-1.5

+

1/—

-I-7S

where

p

la per

cent

«rror.

For

p-0

1

10

'^

-0

0.0932

0.280

Note.

—

Fw

p-0,

expreasion

(11)

beoomes

—

. To evaluate

00

this indefinite

form,

multiply

the numerator

and

the

denominator

by

p

before

puttinf^

p

0. In

the

solution

of

this

and of

the pre>

ceding

problem the

last

term

in

ex

p

r

ess

i

on

(10)

is

neglected on

aeoount

of its

relative

insignificance;

otherwise

it

would

be

neee»>

aary

to solve

a

otnnplicated

cubic

equation,

and

the

results would

be very

little different

from those

obtained above.

Prob.

9.

—

Evaluate expression

(11)

for

the

value

p-0,

accord-

ing

to

the

general

rule

given in the calciilus

for

indeterminate

a|>r»>o..(thefonn^.

Prob. 10.

—

The

required moment

of

inertia

of the

rim

of

a

fly-

wheel

is

9.25

m*.

The

wheel

is

to

run

at

75

rev.

per

min.,

and

its

peripheral

speed must not exceed 25 m. per

sec.

(safety requirement).

CooaiderataoiiBofqMioelimitthewidthof

therimto30cm. Det«^

mine

the necessary

thickness

/

of

the

rim.

Aru.

r

-

3.18

m.;

t

-

16.5

cm.

Hint.

—

Neglect

all the terms

but

the

first in Eq.

(10)

and

calcu-

Ute

/

in

the

first

approximation.

With the

value of I

so

obtainea

calculate the

expression

in the

bracket«, and

again

solve for

t

out-

aide the

brackets. This

will give

the value of

(C

in

a

second

approxi-

mation. If nee

weary, repeat the

process

once

more.

Prob. II.

—

What is

the

moment

of

inertia of a

solid

disk, of

radius

r

and

width

b, about

its geometric

axis

perpendicular

to

r?

.

wvr*

Ahm.

-y.

Prob.

la.

—

Referring to

the

preceding

problem,

divide

the

disk

into

ten concentric

parts,

each

of

which

contributes

the



54

ENGINEERING

MATHEMATICS.

[Chap.

VI.

amount

to the

total moment

of

inertia. Show the

concentric

circles

on

a

drawing to

scale.

Ans.

n

=

r

yf07l

=

0.5623

r;

ri=r

VoT9

=

0.9740 r.

Prob.

13.

—

Let the

concentric

parts,

considered

in

Prob.

12,

be

removed

one

by one,

beginning

with

the

inner

one.

Plot curves

or

tabulate

values showing:

(a) the amount

of reduction in the moment

of

inertia

in

per

cent of the

moment of

inertia

of the

solid disk;

(6)

amount

of material

removed,

in

per

cent of the weight of the

solid

disk;

(c)

the

moment

of inertia

per

unit

volume.

These

curves

or

tables

are

intended

to

show

the advantage of

concentrating

as

much

material

as

possible

near

the

periphery of a

flywheel,

where

the

speed is at its

maximum;

in this

way the largest

possible moment

of

inertia

is obtained

with

the

smallest amount of

material.

Ans.

Number of

cylin-

ders

removed

from

interior.



Cbap.

VI.)

MOMEST

OF

INBHTIA

OF

FLYWUBEI

Solution.

~

Let

4

-

Z

BQN

be

the

angle

of

tnclinattoo

ci the

Qonparaliel sides; then, from

the

triaagleAf.VP,

MP>

NPeoi4,

I

(y

-

6)

-

(r

-

x)

cot

;

whence

y -(6+2rcot4) -2xcot#

(17)

ituting

this

value of

y

into

(16)

gives,

after

integratioo:

/-r}(j6

+

rcot#)(r*-r,«)-0.8oot*(r*-r,»)|,

or,

reducing

to

a

form

oMure

convenient for

raculations.

When

(

-

j

is

small

oom|>ared

with unity,

i

:s

.simplified

to

/-»r*f|j^+0.2cot

A (18a)

The

value

of

cot

#

in

these

expressions

is

determined

from

the

triangle

QiV

A:

eot#-cotZ«gA-^^2l^

(19)

Prob.

ai.

—

Show that

for

a

rectangular

cross section

formula

(18)

becomes

identical

with

(8),

and formula

(18a)

with the

answer

to

Prob. 11.

Prob. 22.

—

How

should

formula

(18)

be modified

for

numerical

calculations

when

t is

small

compared

with r?

An..

/.2WHJ6[.-1.5(i)

+

(f)'-...]

+

rc.f[(l)-2Q'+...]j.

.

.

(20)

Hint.

—

See

solution to

Prob.

5.

Prob.

aj.

—

Apply

formula

(18)

and

(20)

to the

following case:

6

>

20

cm.;

c

->

30

cm.;

(

«

45 cm.;

r

-

3 m.

'

Am.

7-15

m.»

(nearly).

Prob.

24.

—

What

is the radius of

g>'ration, and the average

radius

of the

rim,

in the

pn>ce<iing problem?

Sdidion.

—

According

to Ciuldinus'

theorem

(see

Prob.

26),

the

volume of

the

rim, p

•

5

•

2 w (r

-

Ot

when

N

i-^

t

he

area

of

the

eroai •eetion of

the rim,

and C is the distance

of

the

center

of

gravity

of the

eroM

sectioo

from

tU

outside

edfe

S

V

(Fig.

16).

But

5-^^-^'.

and

f'l'^^f^-

SubsUtutiug

the



68

ENGINEERING

MATHEMATICS.

[Chap.

VI.

numerical

data,

we

find v

=

1.951

cu. m.

Hence,

according

to

Eq.

(12),

k

=\/

^^

=

2.773

m.

The

mean

radius

of

the

rim

V

1.951

45

Tm

=

3.00

'—

=

2.775

m.

Thus, in the

case

under considera-

tion, it

is

accurate enough

for

most

practical purposes

to

calculate

the moment of inertia

by

multiplying the

volume of the

rim by

the square

of its

average

radius,

instead

of

using the

more

compli-

cated

formula

(18).

Note.

—

Expressions

for

t'

used

above

will

be

found

in

various

engineering

handbooks

and pocket books,

also

in

some

works on

calculus and mechanics,

in

the

chapters

on

center

of gravity of

plane

figures. If

the

student

prefers to

calculate

the volume of the

rim

without

the

use

of

Guldinus'

theorem,

or expression

for

t', he may

do so,

as

is

indicated

in the

next

problem. As

a

rule, however,

it

is

recommended that the

student

of

engineering

should

use

all

possible helps

in the form of

auxiliary

relations, tables, etc., when

solving

a

particular

practical

problem, and not to shun

them

simply

because

he did not have

them

in

his course

in

calculus

or

mechanics.

Prob.

25-

—

Determine

the

volume v required

in

the

preceding

problem, without

the use

of

Guldinus'

theorem.

Ans.

„.2„^|(A

+cot*)[l

-

('-)']

-i-eot*[l

-

{^)']\.

Hint.

—

Substitute the

value

of

y

from Eq.

(17)

into the

general

expression

(15),

and

integrate

between

the

limits ro

and

r.

Prob.

26.

—

Prove that

the

volume of a

body

of

revolution

is

equal to

the

area

of

its

cross

section, multiplied by the

circumference

of

the

circle passing

through

the

center

of gravity

of

this area.

This

is

the

well-known

Guldinus'

theorem used in Prob.

24

above.

Proof.

—

According

to

formula

(15),

the volume

of a

body of

revolution

is

..dx

(21)

;

=

27r

A

But,

according to

the

definition

of the

center

of

gravity,

ydx'X

=

Sxo,

(21a)

h

where

Xo

is the

distance

of the

center

of

gravity of

the area

from

the

same axis

from

which

distances

x

are

counted;

in

this

particular

case,

from the

axis

of rotation.

Comparing

Eq.

(21)

and

(21a)

we get: v

=

2

irxo

•

S,

which

proves

the

theorem.



GkAF.

VI-l

MOMENT

OF

INERTIA

OF

FLYWHEELS.

69

Prob. a?*

—

Wliat is the

moment

of

inertia

of a

gpinning top

having a

rim of

circular

eron

aectkm

of radius

a

(Fig. 17)?

Only

an

approximate

solution is

required,

aswiming

the

radius

of

gyrati<m

to

be

equal to

the

average

radius (r

-

a) of

the rim;

an

eiaet

solu-

tion

is

coasidered

in

the next

(xoblem.

Ant.

/-2w»aV(r-a)|l-2(^)-|-(?yj.

.

.

(22)

Prob.

aS.

—

What is the

exact

exp

r

ession for

the moment

of

inotia

required

in

the preceding

probldn?

^

Axhotlautkta

Fig. 17.

—

A

crosB

MCtioo

of

a flywheel with

a

circular

rim.

ScivHon.

—

This

is

a

specific

case

of

intof(ratton

of

expression

(16)

;

instead

of e

x

pres

si

ng

y

as

a

function

of x,

it is

more

oonvenient

in the

case

of a

circle to express both

x

and

y

through

the

central

angle

4

(Fig.

(17).

We

have:

^

•a8in4;x-r-a-aoos4:

dlr

•

a sin

4

<i#.

Substituting

these

values

into

Eq.

(16),

this

ex-

prcMsinn

involves

four integrals whose values are as

follows:

I

sin»#<i#-^;

I

sin*

#

cos

4

<i#

-

0;

I

sin*

cos*

4

d#

-

^

:

j

maf

4

eoif

4

d^

^

0.



60


[Chap.

VI.

So that

7

=

2x2aV2(r-a)jl

-2^^)

+

1.75f^yj. .

.

(23)

It

will

be seen that

for

small values of the

ratio

-

this

expression is

practically identical

with

Eq.

(22).

Prob.

29.

—

Solve the

foregoing

problem

without

introducing

angle

<i>.

Hint.

?/

=

2 a

Vl

—

2*,

where

z

=

~

^

-

a

Prob.

30.

—

At

what

ratio

of

-

does the

error

in

Eq.

(22)

reach

n

per

cent, as compared

with the

correct

solution

(23)?

Ans

For

instance,

when

n

=

1,

-

=0.104.

r

Prob.

31.

—

A

disk

of variable

width

is

to be

designed

so

that

each

infinitesimal

concentric

layer

of

material

should

contribute

the

same amount

to

the

total

moment

of inertia.

What

is the

shape of

the profile of the

disk?

Ans. yx^

=

Constant

[according

to

Eq.

(16)].

^

Prob.

32.

—

Referring

to the

preceding

problem,

draw

to

scale

the cross section

of a

solid

steel

disk, which

must

satisfy

the following

conditions: The

disk is to run

at a speed

of

1500

rev. per min.,

the

peripheral

speed

not

to

exceed

80

m.

per

sec; the

diameter

of

the shaft

is 20

cm. ;

the

total energy

of the disk

must

be 8100

kg-m.

(without

shaft).

Specific

gravity

of

steel is

7.86.

Ans. Diameter of the disk

is

1.018 m.; width at

the periphery

=

1.95

mm.;

width

at

the

shaft

=

25.6

cm.

Prob.

33.

—

What

is the

general

expression

for

the

moment

of

inertia

of

each

arm (or

spoke) of

a

flywheel

(Fig.

15)?

Ans.

According to

Eq.

(6)

sdx,

(24)

where

s is

the variable

cross

section

of

the arm,

and

must

be given

as

a

function

of

x.



IT

bap.

VI.J

MOM

By

T

OF INERTIA

OF

FLYWHBBLS.

61

Prob.

34.—

Integrate

expreasion

(24)

in

the

first

approxunation,

by awHiming

the

aro«

aeetioo of the aim

constant

and

equal

to Ha

mean

value

««.

Am.

/-|«.(r,»-r.')

(25)

Prob.

35.

—

Prove that, for

a rough approximation,

the

mmnent

of

inertia of

the arms

can be

taken

into

account

by aaauming

0De>

third

of their wei|^t concentrated

in

the

rim.

Solution.

—

Expreencm

(25)

can

be

reiKesented

in

the

form

/

-

J

«•

•

(r.

-

r.)

(r,»

+ r^i

+

r,«).

But

Sm

•

(r«

-

rO b

approximately

equal to the

volume

V

of

the

arm,

so

that

The

two

ia«i

terms within

the brackets

are

small as

compared with

unity;

on

the other hand,

r«

is

somewhat

smaller than

the

radius

of

gyration

k of the

rim.

Assunoing

that the

expression

within

the

brackets

makes

up

for

this

diflferenoe,

we

get

approximately

/

-

i

Vk*

(26)

Comparing

this with Eq.

(12)

proves the

prqxMition

and

justifies

the

simple

practical

formula

/loul

-

{Wrim

+

ifWwrm)

rj

(27)

which

is

much

used in the preliminary design of

flywheels.

In

this

formula

W

denotes

wdght,

/

is the number

of

arms,

and

r.

is

the

average

radius

<rf

the rim.

Prob. 36.

—

Calculate the

moment of inertia

of an

arm

more

aecuratdy than

in

Prob.

33;

namely, assume that the cross

seo-

tKNi

«

varies

according

to the

straight-line law,

between

its

extreme

values

«•

and

$t.

Ans.

The

variable croas

section

b represented by

M'A-Bz

(28)

and

Eq.

(24)

gives

,.^ifVzjV}_B( V^^

....

(29)

where the

constants

A

and

B

are determined

by the

conditions

:::':*:

'^^

Prob.

37.

—

Obtain

a

still

closer value for the

moment oi

inertia

of

an

arm,

by

cooaideriDg

the

arms

as

ttraight

truncated

oonea.



62

ENGINEERING

MATHEMATICS.

[Chap.

VI.

Solution.

—

Let

the fictitious

vertex

of the

cone

be

at

a

distance

R

from the

axis

of the flywheel.

Then,

for a cross

section

s

at a

distance

x

from

the

axis,

we

have

So

(R-nr'

••••••

^^^)

Substituting

this value

of

s

into Eq.

(24)

gives

after

integration

and

reduction

In

this^orm

the expression is

convenient

for

numerical

calculations.

The

correction

terms,

of

comparatively

small

magnitude,

are

directly

apparent, and

the

formula

resembles

the

approximate

expressions

deduced in Probs.

33 and

34.

The

distance

R

is

deter-

mined

from Eq.

(31),

by applying

it

to

a;

=

n;

we have,

then.

«l

_

(fi

-

TrY

So

(R

-

rof

whence

ro

R

Al4«

L

(33)

When

the

arms are

cylindrical in

shape,

Si

=

So;

jR

= oo

;

^

=

0;

R

and Eq.

(32)

becomes identical

with

(25),

as is

to

be

expected.

Prob.

38.

—

A

flywheel

has a diameter

of

4.8

m.,

and

the radial

thickness

of

its rim

is

38 cm.

The

wheel

is provided

with

12

arms

of

elliptical

cross section;

the

principal

dimensions

of

the

cross

section

at

the

rim

are

4.8

by

8 cm., and

at the hub 6

by 10

cm.

The

arms

are

1.77 m. long.

Calculate the

moment

of

inertia

of

one

arm,

according

to the three

approximations

considered

in

the

previous

problems.

Ans.

/=

0.0106 m.*

according

to

formula

(25)

0.0096 m.*

according

to

formula

(29);

0.0095

m.

according

to

formula

(32).



Cbaf.

VIl

MOMENT OF

INBRTIA

OF

FLYWHEELS.

68

Note.

—

Dr.

F.

R.

Shvpe

has called the

autbor's

attention

to

the

following

useful

fonnula

for ralniiating

the

moment at

inertia

ot a

rim.

Let

the distance x between the axis

of

rotation and

a

point oo

the

rim

be

represented as

a

sum

of

two distanees,

xt

+

X|. Here

x*

is

the

distance between

the

axis of

rotation

and

a parallel

axis

pass-

ing

throui^

the

center

of gravity

of

the

cross

section

of the rim;

this

axis

will

be refored to

betow

as the

axis BB;

xi

is

the

<fistanoe

fron

the

axis

BB

to the

point

undareoDsid««tion. Besides,

in

£q.

(6)

we

can

put

dp-

2wx

-dA,

where

<fil

is an infinitesimal

element

of croos

seetioo

at

the

point

under

consideration.

8ub>

stititting

into Eq.

(6),

we get

/

-

Czwiz.

+x,)»di4

-

2w^xM

+

3x,* fx^A

•^3x,fit*dA+Cxt^Al.

The

first

oi

the three

integrals

is

equal

to

sero,

bong

the static

moment

of

the

crosB

section

with

respect

to

the

axis

BB,

passing

through

the

center

of

gravity.

We

have, thtfef<Mre,

/

-2»}x.».4

+

3x,K,+

A'.j.

Here

Kt

is

the

moment

of

inortia

of

the cross

section

with respert to

the

axis

BB.

For

usual

forms

of cross

section,

expressions

for

A'l

are

given

in

engineering

pocket books.

A',, for lack

of

a

better

name,

can

be

called the

cvbie

moment

of

the

cross

section, with

re-

spect to

the

axis

BB.

If

the

cross section

is

symnietncal

with

req>ect to

the

axis

BB,

A'a

-

0,

because to each positive value

of

Xi»dA

oorreqwnds

an

identical

negative value.

Thus,

for

qrm-

metrical

cross

sectiras,

we

hsN-e

simply

/-2rt,jx.M

+

3A,j.

For

unqrmmetrical

cross

sections

the

value

of K, has to

be

calcu-

lated,

so that the

method

does

not

possess

any

advantage over that

given

in the

text

above.

Hofwever,

with

moderate

unsymmctry

in

theshape,

the

value

of

A.

is very

small

as compared

with

the

two

other

terms

in the

expression

for /. It

can,

therefore, be

neglected,

or

calculated

under

simplifying

assumptions



LIST

OF REFERENCE

WORKS

ON

MATHEMATICS.

(See also

the

books

recommended

in

the

appendix.)

TRIGONOMETRY.

Author.

Publisher.

AsHTON

&

Marsh

Chas.

Scribner's

Sons.

MoRiTZ

John

Wiley

&

Sons,

Inc.

Jones

Geo. W.

Jones

(Ithaca).

HoBsoN

Cambridge University

Press.

Palmer

&

Leigh v

McGraw-Hill.

ANALYTIC

GEOMETRY

AND

CALCULUS.

Woods

&

Bailey

Ginn

&

Co.

ANALYTIC

GEOMETRY.

Tanner & Allen

(revised

edition)

Amer.

Bk. Co.

Smith

&

Gale

(larger edition) Ginn & Co,

Fine

&

Thompson

The

Macmillan

Co,

C.

Smith The

Macmillan

Co.

Salmon Longmans. Green

&

Co,

Ashton

Chas.

Scribner's

Sons.

Wilson & Tracy

D.

C. Heath

& Co.

Phillips John

Wiley

&

Sons,

Inc.

DIFFERENTIAL

AND INTEGRAL

CALCULUS.

{Both

in

one

volume.)

Snyder

&

Hutchinson Amer.

Bk.

Co,

Granville Ginn

& Co,

Murray

Longmans,

Green

&

Co,

Osgood

The

Macmillan

Co.

Echols

•

Henry

Holt

&

Co.

Lamb

University

Press,

Greenhill

The

Macmillan Co,

PuiLLii'S

.John

Wiley

&

Sons,

Inc,

04



LIST

nh'

RFFBRBSCB

WORKS

ON

MATHKMATICS.

65

DIFFERENTIAL

CALCULUS.

AiTHOR.

PrnusiiKu.

McMahon a

Sntdcb

Amrr. Bk.

Co.

Btbrly

t^'inn

A

Co.

[EowAsos...

The

MMtnillan

Ca

IVfiMAAAuaos

.

Longmana,

Grem

A

Co.

Pbiluk

John Wiley

ft

Son*,

Inc.

INTEGRAL

CALCULUS.

Aiiicr.

Bk. Co.

(Jinn

&.

Co

EdwaBDS.

.

.

Thi' Macinilian

Co.

WlLUAMBON.

i/>nKnianii,

(trrt'n

&

C<i.

ToDRUMTER

- The

MacmilUtn

Co.

Philups

John

Wiley

ft

Sons, Inc.

HYPERBOUC

FUNCTIONS.

McMahon

John

Wiley

ft

Sons,

Inc.

.M \THEiiATirAi.

Tabi.ks .

Smithsonian

Inst.

TABLES OF

INTEGRALS.

B.

O.

PURCB

Cinn

ft

Co.

HuosON ft

LiPKA

(Reprinted

from

Kngineerx'

Manual)

lohn

Wiley

ft Sona,

Inr.

Peibcb ft

Cabvbr

McGraw-Hill.

MISCELLANEOUS.

Mkrum

AN-

Woodward.

Mathematical

Monographs.

John

Wiley

ft

Sonn.

Inc.

History of

Modem

Mathematics,

by

David

Kugrnc iSmith.

2.

Synthetic

Projective

Geometry,

by George

Bruce

Uakted

Determinants,

by

Laenas

Gilford

Weld.

Hyperbolic

Functions,

by

James

McMahon.

Harmonic

Funetioas,

by

William

E.

Byerly.

Graasman's

8p«oe

Analynt,

by

Edward

W.

Hyde.

Probability

and

Theory

oC

Errors,

by

Robert

8. Woodward.

Vector

Aiudysis

and

Qufttornkwa, by Alexander

Maefariane*

Differential

EquaiioM,

by

WtUiMn Woohe

y

JohnMo.

No.

10.

Solution

o(

Equations, by

Mannfield MerrUnan.

No.

11. PuDetiona

oT

a

Complex

Variable,

by Thomas

8. Fiske.

No.

12. The

Theory

of

Relativity,

by

Robert D.

Carmiefaad.

No.

13.

The

Theory

ct

Numbets, by

Robert

D.

OamkbaeL

No.

14. Alfebraic

Invariants,

by

Leonard E.

Diekaon.

No.



66

LIST

OF

REFERENCE

WORKS

ON

MATHEMATICS.

No. 15.

Mortality Laws

and Statistics, by

Robert

Henderson.

No.

16.

Diophantine

Analysis,

by

Robert

D.

Carraichael.

No.

17.

Ten

British

Mathematicians, by Alexander

Macfarlane.

No.

18.

Elliptic Integrals,

by

Harris

Hancock.

Problems

in Calculus, Leib

Ginn

&

Co.

Review

of Algebra, Romeyn

Henry

Amer.

Bk.

Co.

College

Algebra, Wilczynski

&

Slaught

Allyn

&

Bacon.

Manual

of

Mathematics

(Reprinted from

Engineer's

Manual),

Hudson

&

Lipka John Wiley

&

Sons,

Inc.

College

Algebra,

Dickson

John Wiley

& Sons,

Inc.



(H

1CEFEUL.NCE

BOOKS

ON MACHINE

DESIGN.

Bacb..

Bbmjamin

.

JONBB

Kimball

A

Baki

Smith

A

Marx.

Spoonbs.

Uhww

.

JuliuA

Springer,

iionry

Holt

A.

Co.

I'lhn

Wilpy

A

Sons,

Inc.

John

Wiley

ft

Sooa, Inc.

John Wiley

ft

Sons,

Ine.

LongnuuM, Green

ft

Co.

Longmans,

Green

ft Co.

«7



70

APPENDIX.

6. Application

of

mathematics to mechanics and

physics.

Go over

your

textbooks

of

physics

and mechanics,

and

select

topics which

involve

a considerable

use

of

mathematics.

Solve

the

most

important

problems in

these

topics,

paying

the

principal attention

to

the

mathe-

matical

methods

employed,

rather

than to

the

physical

side of

the

questions.

This

will

be

a

valuable

preparation for

the

application of

mathematics

to engineering.

GENERAL

REMARKS

1.

Do

not

try to

cover too

much

in

the

above

outlined

program.

Select

a

topic in which you

are particularly

interested,

or

in

which

you

feel

particularly

deficient,

and go

over

it carefully, before beginning

the

next topic. One subject in

mathematics studied

thoroughly

will

give you more insight

into the

general mathematical method,

than

a

large

field

covered in

a

superficial way

and

only

half

understood.

Many

troubles

in

the understanding of

the

calculus have their origin

in

insufficient

preparation

in

elementary

mathematics.

Therefore,

it

is

advisable

to

review

the

principal parts of

algebra

and

trigonometry

before reviewing

the

calculus.

2.

Having

selected

a

topic, begin its

study by solving

a

consider-

able

number

of

problems, so

as

to acquire a fluency

in

applications.

Refer back

to theory

only in

so

far

as you

find

difficulties

in

the

solu-

tion of

the

problems,

or

in understanding

the

reasons

for

certain rela-

tions.

3.

A

convenient

reference

book for

all

the

above-mentioned

topics

in

elementary

mathematics

and

in calculus

is

the Mathematical

Hand-

book by

E.

P.

Seaver (McGraw-Hill,

$2.50).

No proofs

are

given

in

this

book,

but

only

the

principal

formulae

and

rules,

arranged

in

a way

convenient

for ready

reference.

A

somewhat larger

book

is

that

by

J.

Claudel,

—

Handbook of

Mathematics (McGraw-Hill,

$3.50).

4.

Those

particularly interested

in application of

mathematics

to

en-

gineering

will

find

a

large

number of

engineering

problems

with

solu-

tions

in

the

following

works:

F.

M.

Saxelby.

A

Course in

Practical Mathematics

(Longmans,

$2.25).

See

in

particular the

examination

papers, beginning on

p.

396.

This book

is

particularly

recommended for

home

study.

F.

Castle.

A Manual of

Practical

Mathematics (Macmillan,

$1.50).

R.

G.

Blaine.

The Calculus

and

its Applications (Van

Nostrand,

$L50).

John Perry.

Calculus

for

Engineers

(Longmans,

$2.50).

Chas.

P.

Steinmetz.

Engineering

Mathematics

(McGraw-Hill,

$3.00).

5.

Every

year some

seniors

have

difficulty in

grasping

the

theoreti-

cal

subjects

in engineering,

simply

because

of

insufficient

preparation



APPFSmX

1

in

malhomatic *.

Tliey

f;iii in ^plt^

ms

(uiincnl

work.,

IktuU ^- t)»«-n-

is

DO

liiHO

during

tin-

colltnc

yiar

tu

dt-vutt-

to

tin-

ri-\u-w

of

in;itlifnmtir«.

It

18 therefore urgently

reoommended

tlut tbooe

who

feel

deficient

in

nuitheuatieit

go

uver

the subject

during

tiie

summer

bcfwe

tbe

aei^r

year.

Not

only

will

poor

prepaimtioa

In

matbemAtiet

be

eonridored

no

excuse

for

unsftttsfaotory

work in englneerins

subjects,

but, on tlie

|«oatr»ry,

it

may

serious^

interfere

with

graduation.



PLEASE

DO

NOT

REMOVE

i

CARDS

OR

SLIPS

FROM

THIS

POCKET

UNIVERSITY

OF

TORONTO

LIBRARY

TA

330

K3

1917

PT.l

C.l

EN6I

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